Beta Decay:
0^131 I → -1^0 e + 53^131 Xe
0^32 P → 15^32 S + -1^0 e +
24^11 Na → 0^24 Mg + 11^e +
0^241 Pu → 94^241 Am + -1^0 e +
The decay of the given nuclei through the emission of a beta particle can be represented by the following nuclear equations:
0^131 I → -1^0 e + 53^131 Xe
In this equation, the nucleus of iodine-131 (131 I) undergoes beta decay, resulting in the emission of a beta particle (e-) and the formation of xenon-131 (131 Xe). The atomic number of iodine decreases by 1 (from 53 to 52), while the mass number remains the same (131) since the beta particle carries negligible mass.
0^32 P → 15^32 S + -1^0 e +
Phosphorus-32 (32 P) undergoes beta decay, resulting in the emission of a beta particle (e-) and the formation of sulfur-32 (32 S). The atomic number of phosphorus increases by 1 (from 15 to 16) due to the conversion of a neutron into a proton.
24^11 Na → 0^24 Mg + 11^e +
Sodium-24 (24 Na) undergoes beta decay, resulting in the emission of a beta particle (e+) and the formation of magnesium-24 (24 Mg). The atomic number of sodium decreases by 1 (from 11 to 10) as a neutron is converted into a proton.
0^241 Pu → 94^241 Am + -1^0 e +
Plutonium-241 (241 Pu) undergoes beta decay, resulting in the emission of a beta particle (e-) and the formation of americium-241 (241 Am). The atomic number of plutonium increases by 1 (from 94 to 95) due to the conversion of a neutron into a proton.
It is important to note that the specific isotopes produced in the decay reactions may vary depending on the initial nucleus and its specific decay pathway. The selected isotopes in the equations above are based on the information provided.
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12. A hiker walks \( 300 \mathrm{ft} \mathbf{1 5} \) degrees north of west and \( 0.7 \) km northeast. Calculate the magnitude and direction of the hiker's total displacement.
The hiker's total displacement(HTD) is approximately 77.63 ft at an angle of 0.365 degrees north of west.
The displacement of the hiker can be calculated using Pythagoras's Theorem(PT) and trigonometry (Tgy) . To do so, we need to break the displacement into its x- and y-components. Let's start with the x-component of the displacement(d): It's the component pointing in the north direction. Since the hiker is walking 15 degrees north of west, that means they are walking at an angle of 75 degrees with respect to north: (90 degrees - 15 degrees). Using trigonometry, we can find that the x-component is equal to:$$\begin{aligned}x &= 300 \cos 75^\circ\\&= 300 \cdot 0.258819\ldots\\&= 77.65 \mathrm {ft}\end{aligned}.
Now let's find the y-component of the D. This component points in the northeast direction, which means it is 45 degrees away from both north and east. Using trigonometry again, we can find that the y-component is equal to:$$\begin{aligned}y &= 0.7 \cos 45^\circ\\&= 0.7 \cdot 0.707106\ldots\\&= 0.495 \mathrm{km}\end{aligned}$$Now we can use PT to find the magnitude of the displacement:\begin{aligned}d &= \sqrt{x^2 + y^2}\\&= \sqrt{(77.65 \mathrm{ft})^2 + (0.495 \mathrm{km})^2}\\&= \sqrt{6025.9125 + 0.245025}\\&= \sqrt{6026.157525}\\&\ approx 77.63 \mathrm{ft}\end{aligned}$$Finally, we can use Tgy again to find the direction of the displacement. This is given by the angle that the displacement vector(Dv) makes with respect to north. We can find this angle using:$$\begin{aligned}\theta &= \tan^{-1}\left(\frac{y}{x}\right)\\&= \tan^{-1}\left(frac{0.495 mathrm {km}{77.65 \mathrm {ft}\right)\\&= \tan^{-1}(0.006372ldots) & approx 0.365^circ\end{aligned}.
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In a Cartesian coordinate system (x,y,z) between the two points P1= (1 cm, 2 cm, 1 cm) and P2= (4 cm, 2 cm, 6 cm) there is an electrical field which directs along the connection line from P1 to P2 at any point. The magnitude of the electrical field increases like 5Vcm3/s2, where s is the distance from point P1. Calculate the electrical potential at a distance of 2 cm from Point P1 when the electrical Potential at a distance of 4 cm from Point P1 is zero
The electrical potential at a distance of 2 cm from Point P1 when the electrical potential at a distance of 4 cm from Point P1 is zero is 1.25 V.
Given, Two points P1 and P2 in Cartesian coordinate system (x,y,z) as shown below: P1= (1 cm, 2 cm, 1 cm) and P2= (4 cm, 2 cm, 6 cm)
Electric field, E increases like 5Vcm3/s2, where s is the distance from point P1.
Distance between P1 and P2 = 5 cm
The direction of electrical field is along the connection line from P1 to P2 at any point. The potential difference between P1 and P2 is the negative integral of the electric field over the distance from P1 to P2.V = - ∫E.ds, where E = 5Vcm3/s2 and s = distance from P1 to P2∴ V = - 5 ∫ds/s3 = 5/s + C
Where C is a constant of integration.
When V = 0 at a distance of 4 cm from P1, the constant of integration, C can be calculated as follows: 0 = 5/4 + C => C = -5/4
Therefore, V = 5/s - 5/4
At a distance of 2 cm from P1, s = 2 cm∴ V = 5/2 - 5/4 = 5/4 V = 1.25 V
Therefore, the electrical potential at a distance of 2 cm from Point P1 when the electrical potential at a distance of 4 cm from Point P1 is zero is 1.25 V.
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0. A uniform beam fixed at one end and simply supported at the other is having transverse vibrations. Derive suitable expression for its frequency equation.
The beam is subjected to transverse vibrations. Transverse vibrations occur when a beam vibrates in the direction perpendicular to its axis. The frequency of a beam that is uniform in cross-section is directly proportional to the square root of its stiffness or elasticity.
The following is the formula for calculating the frequency of a beam under transverse vibrations:F = (1/2π) × √(EI/mL²)Where F is the natural frequency, E is the elastic modulus of the material, I is the second moment of area, m is the mass of the beam, and L is the length of the beam.
Let the beam be fixed at one end and simply supported at the other, as shown in the following diagram. As a result, the beam's effective length is L, and its effective mass is m. We can use the equation above to calculate the natural frequency of the beam in this configuration.
In this case, the frequency of the beam's transverse vibration is given by the following equation:F = (1/2π) × √(3EI/mL³)This is the expression we're looking for.
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When a voltage-gated sodium ion channel opens in a cell membrane, Na+ ions flow through at the rate of 1.8 x 10 ions/s What is the current through the channel? Express your answer with the appropriate units.
The current through a channel when a voltage-gated sodium ion channel opens in a cell membrane can be calculated using the formula `I = q/t`, where q is the charge that passes through the channel and t is the time taken for that charge to pass through the channel.
We can use the formula `q = n * e`, where n is the number of ions passing through the channel and e is the charge on a single ion.
The given rate of Na+ ions passing through the channel is `1.8 x 10^10 ions/s`. Therefore, the number of ions passing through the channel in time t is `n = (1.8 x 10^10 ions/s) * t`.
The charge on a single ion is `1.6 x 10^-19 C`.
Therefore, the total charge passing through the channel in time t is `q = n * e
= (1.8 x 10 ions/s) * t * (1.6 x 10-19 C/ion)`.
Substituting these values in the formula `I = q/t`, we get: `I = [(1.8 x 10ions/s) * t * (1.6 x 10 C/ion)] / t
= 2.9 x 10 A`.
Therefore, the current through the channel is `2.9 x 10 A`. The appropriate units for current are amperes, which is represented by the symbol A.
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45. A
60
Co source is labeled 6.1mCi, but its present activity is found to be 2.03×10
7
Bq. (a) What is the present activity in mCi ? mCi. (b) How long ago in years did it actually have a 4.00−mCi activity? years. 45. A
60
Co source is labeled 6.1mCi, but its present activity is found to be 2.03×10
7
Bq. (a) What is the present activity in mCi ? mCi. (b) How long ago in years did it actually have a 4.00-mCi adtivity? years.
The present activity in mCi is 0.610 mCi. The 60Co actually had a 4.00-mCi activity 20.8 years ago.
Given, Activity of 60Co = 2.03 × 107 Bq = 6.1 mCi
(a) We have to find the present activity in mCi.
Activity = 6.1 mCi = 6.1 × 10−3 Ci = 6.1 × 10−3 × 3.7 × 1010 Bq = 22.57 × 106 Bq = 2.257 × 107 Bq
Present activity in mCi = 2.257 × 107/3.7 × 1010= 0.610 mCi
Therefore, the present activity in mCi is 0.610 mCi.
(b) We have to find how long ago in years did it actually have a 4.00-mCi activity.
Activity of 60Co = 4.00 mCi = 4.00 × 10−3 Ci = 4.00 × 10−3 × 3.7 × 1010 Bq = 14.8 × 106 Bq
Let 't' be the time for which it actually had a 4.00-mCi activity.
Hence, the initial activity (A0) = Activity of 60Co at present (A) = 2.03 × 107 Bq.
The activity of radioactive substance is given by the relation, A = A0e−λt, where, λ is the decay constant, which can be calculated as follows: A0 = A = 2.03 × 107 Bq = A0e−λtλ = -ln(2)/T1/2 = -ln(2)/5.27 = 0.1314/day
Putting the values of λ, A0, and A in the above relation, 2.03 × 107 = A0e−0.1314tA0 = 2.03 × 107 /e−0.1314t= 2.03 × 107 / (1/2.718)0.1314t= 2.03 × 107 × 2.7180.1314t= 5.51 × 107t= (1/0.1314) ln (5.51 × 107 / 2.03 × 107)t = 20.8 years
Therefore, the 60Co actually had a 4.00-mCi activity 20.8 years ago.
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#3 Energy and dynamics of Bungee Jumping Bungee Jumping is a combination of a free fall (upper part) and a harmonic oscillation (lower part). You are asked to specify a proper elastic rope (that works like a spring) for Bungee Jumping off the Emmerich Rhine Bridge. The height above the Rhine level from the top is 100 m and the mass of the person jumping is 85 kg. You would like to have half of the distance in free fall and the other half moving while the elastic rope is stretched (which perfectly follow's Hooke's law). Just at the Rhine level you are supposed to have zero velocity Calculate the required spring constant of the elastic rope. What is the velocity just after the first half of the height of free falling? How long does it take to go half the way? Sketch an acceleration-time, a velocity-time, and a displacement-time diagram. Consider everything you have learned about proper diagrams! Calculate/mark all axis intersections quantitatively, curvatures (if there are any) qualitatively. rope length unstretched spring constant kinetic energy alier of the rope Talling hallway time until falling (diagrams) points h all way (see above) 14 1 4 715 /28
Energy and Dynamics of Bungee Jumping Bungee jumping is an interesting sport that combines free-fall and harmonic oscillation. An elastic rope is used in bungee jumping.
It functions as a spring that obeys Hooke's law to ensure that the jumper has a safe landing. In this question, we are supposed to determine the spring constant of the elastic rope that will be used for a bungee jump off the Emmerich Rhine Bridge.
We are also expected to calculate the velocity just after the first half of the height of free-falling, the time taken to go haf the way and sketch an acceleration-time, a velocity-time, and a displacement-time diagram. To do this, we will use the formulas of energy and dynamics.
Below is the solution: Given information: Height above the Rhine level from the top = h = 100 m Mass of the person jumping = m = 85 kgWe want the first half of the jump to be free fall and the second half to be stretching of the rope. Since half of the total height is in free fall, the height in free fall (upper part) will be h/2 = 50 m.
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The sun's apparent location in the sky east or west of true south is called: Azimuth Altitude Irradiance GPS location Question 28 (1 point) Solar window Refers to: The amount of sun that comes through
The sun's apparent location in the sky east or west of true south is called Azimuth. Azimuth is the angular distance of the sun measured clockwise from the North to the position where the sun is at a particular time in the sky, which is east or west of true south.Referring to solar energy,
the Solar window is defined as the period when a given area receives enough sunlight to make solar energy generation economically feasible. This refers to the amount of sun that comes through and is required for the solar panels to produce enough energy to justify the investment.Therefore, the sun's apparent location in the sky east or west of true south is called Azimuth, and the Solar window is referred to as the amount of sun that comes through, needed for solar panels to produce enough energy to justify the investment.
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When your urinary bladder is full, the bladder pressure can reach up to 60 mm H2O. a Assuming that there is no height difference between your urinary bladder and where your urine comes out, calculate the speed at which your urine comes out. The density of urine is 1030 kg/m3 . b If the diameter of a urethra is 6 mm, estimate the volume flow rate of urine as it comes out in units of liters per second. If a full bladder constitutes 500 mL of urine, how long will it take you to remove all of the urine from your bladder? d Is the answer in c a realistic time for peeing? What should be added to make it more realistic? с
a)The speed at which urine comes out can be calculated using Bernoulli's equation, which relates the pressure of a fluid to its velocity. The equation is:P + (1/2)ρv² = constant
Where P is the pressure of the fluid, ρ is the density of the fluid, v is the velocity of the fluid, and the constant is the same at all points along the streamline. The constant can be neglected because the height difference is negligible. Therefore, at the bladder, P = 60 mmH2O (convert to Pa) and
ρ = 1030 kg/m³:60 mm H2O
= 60/1000 * 9.81 Pa
= 0.5886 PaThus,P + (1/2)ρv²
= 0.5886 PaRearranging this equation to solve for v gives:v = sqrt(2P/ρ)
= sqrt(2*0.5886/1030)
= 0.033 m/s
Answer: 0.033 m/s
b)The volume flow rate of urine can be calculated using the equation:Q = Avwhere Q is the volume flow rate, A is the cross-sectional area of the urethra, and v is the velocity of the urine found in part (a).
The diameter of the urethra is 6 mm, so the radius is 3 mm = 0.003 m:Area = πr²
= π(0.003)²
= 2.827E-5 m²
The volume flow rate is then:Q = Av = (2.827E-5 m²)(0.033 m/s)
= 9.32E-7 m³/s
To convert to L/s, divide by 1000:Q = 9.32E-7 m³/s ÷ 1000
= 9.32E-10 L/s
Answer: 9.32E-10 L/sc)If the bladder holds 500 mL of urine, it will take:Time = Volume flow rate⁻¹
= (500 mL) / (9.32E-10 L/s)
= 5.36E8 s (approx.)
Answer: 5.36E8 s, which is not a realistic time for peeing.
d)The answer in (c) is not a realistic time for peeing because it is several years. The units should be changed to minutes or seconds to make it more realistic. To make it more realistic, the person's rate of urine production should be taken into account. Most people produce urine at a rate of about 1-2 L per day, or 0.7-1.4 mL/min. Therefore, if a person has a full bladder, they should be able to empty it in less than a minute, assuming normal bladder function.
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Masses of 300g and 350g are suspended at the ends of a cord
passive over frictionless pulley. Find the distance the masses will
travel from rest position at the end of 2 seconds.
Given masses of 300g and 350g suspended at the ends of a cord, passed over a frictionless pulley, we need to find the distance traveled by the masses from rest position at the end of 2 seconds. Let's first use the formula of acceleration with mass:
m = F / a where,
m = mass,
F = force, and,
a = acceleration.
In the above equation, we will also substitute force with weight, which is given by We will also find out the total mass and the net force acting on it. The total mass is given by,
m = m1 + m2
= 300 g + 350 g
= 650 g
= 0.65 kg
The net force is given by, [tex]Fnet = F1 - F2[/tex] where, F1 is the force due to mass 1, and, F2 is the force due to mass 2.The weight of mass 1 is given by,
W1 = m1g
= 0.3 kg × 9.8 m/s²
= 2.94 N
The weight of mass 2 is given by,
W2 = m2g
= 0.35 kg × 9.8 m/s²
= 3.43 N
The formula is given by,
s = ut + 0.5 at²where,
s = distance,
u = initial velocity = 0 m/s
t = time = 2 seconds
a = acceleration = 0 m/s² (as calculated above)
s = 0 × 2 + 0.5 × 0 × (2)²s
= 0 m
The distance traveled by the masses from rest position at the end of 2 seconds is zero.
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1. Calculate the temperature reading in Celsius scale if its value is five times than that in Fahrenheit scale. 2. A mild steel is 400 mm long at 18 °C. The coefficient of linear expansion for steel is 11 x 10-6/ K. Calculate the increase in length and the final length when heated to 90 °C.
Therefore, the increase in length is 0.03168 mm and the final length when heated to 90 °C is 400.03168 mm.1. To calculate the temperature reading in Celsius scale if its value is five times than that in Fahrenheit scale, we can use the formula,F = (9/5)C + 32Here, we have to find the temperature in Celsius scale when it's five times than that in Fahrenheit scale. So, let's assume the temperature in Fahrenheit scale to be F, then the temperature in Celsius scale will be C, and we can write: F = 5CUsing this in the above equation, we get:5C = (9/5)C + 32(9/5)C - 5C = 32(4/5)C = 32C = 32 x (5/4)C = 40Therefore, the temperature reading in Celsius scale is 40 °C.2.
We are given the following details:Mild steel is 400 mm long at 18 °CCoefficient of linear expansion for steel is 11 x 10^-6/KWe have to find the increase in length and the final length when heated to 90 °C.The increase in length is given by the formula:ΔL = αLΔTwhere α is the coefficient of linear expansion, L is the original length, and ΔT is the change in temperature.
Substituting the values, we get:ΔL = (11 x 10^-6/K) x (400 mm) x (90 °C - 18 °C)ΔL = (11 x 10^-6/K) x (400 mm) x (72 °C)ΔL = 0.03168 mmFinal length = Original length + Increase in length= 400 mm + 0.03168 mm= 400.03168 mm
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(a) Why beta ()decay occurs in spite of huge positive charge in the nucleus? When control rod is required in the nuclear power plant? What are the advantages and limitations of nuclear fusion reactor?
(b) What is the power output of a reactor fueled by uranium-235 if it takes 30 days to use up 2 Kg of fuel and if each fission gives 198 MeV of energy?
A) The atom's nucleus is made up of positively charged protons and uncharged neutrons, which are held together by the strong nuclear force. Limitations of nuclear fusion reactors: Nuclear fusion is not yet commercially viable, and the technology is still in development. B) Power output = 1.28 × 10^13 J / 2592000 seconds = 4.94 × 10^6 watts (approx).
(a) Beta () decay occurs in spite of a huge positive charge in the nucleus due to the following reasons:
The atom's nucleus is made up of positively charged protons and uncharged neutrons, which are held together by the strong nuclear force.
The repulsion between the protons is counteracted by this force.
However, when a neutron in the nucleus transforms into a proton by emitting a beta particle, the number of protons in the nucleus increases.
This raises the repulsion between the positively charged protons, making it unstable.
As a result, the nucleus emits a beta particle to maintain stability and attain a lower energy state. It happens when the ratio of neutrons to protons in the nucleus is imbalanced.
Control rods in nuclear power plants are used to control the rate of fission chain reactions and regulate the energy generated by a nuclear reactor.
The control rods are inserted or removed from the reactor core to absorb or slow down neutrons, which slows down the reaction and regulates the energy produced. In nuclear reactors, the speed of the reaction must be controlled because a fast reaction produces too much energy, causing the reactor to overheat and leading to an explosion.
Advantages of nuclear fusion reactors:
Nuclear fusion is a safe and environmentally friendly energy source that produces no greenhouse gases and has minimal radioactive waste.
Nuclear fusion does not produce nuclear waste that is difficult to dispose of.
Nuclear fusion can generate large amounts of energy in a small space.
Nuclear fusion requires only a small amount of fuel to produce a large amount of energy.
Limitations of nuclear fusion reactors:
Nuclear fusion requires extremely high temperatures and pressures, making it difficult to achieve and sustain.
Nuclear fusion is not yet commercially viable, and the technology is still in development.
It is expensive to construct and maintain a nuclear fusion reactor.
(b)The power output of a reactor fueled by uranium-235 is 1.28 × 10^13 J if it takes 30 days to use up 2 Kg of fuel and if each fission gives 198 MeV of energy.
Power output = (Total energy released) / (Time)Total energy released = (mass of fuel used) × (energy released per fission)mass of fuel used
= 2 kg × 1000
= 2000 g
Energy released per fission:
= 198 MeV
= 3.168 × 10^-11 J
Using the above formula we get:
Power output = 1.28 × 10^13 J / 2592000 seconds = 4.94 × 10^6 watts (approx).
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Listen An infinitely long cylindrical shell extending between r = 1 m and r = 3 m contains a uniform charge density pv0. Apply Gauss's law to find D in the region r > 3m. (a) Setup equations (b) Show work (c) Final answer
a) Gauss's law can be expressed in integral form as shown below:∫E·dA = Qenc/ε₀ ; b) D in the region r > 3m can be found using the relation D = ε₀E ; c) the final answer is: D = 4 ρv₀r .
(a) Setup equations: We have a cylindrical shell with uniform charge density, ρv₀ .Gauss's law relates the flux of the electric field over a closed surface with the charge enclosed within the surface. Using Gauss's law, the electric field can be found by integrating over a closed surface containing the charge.
Gauss's law can be expressed in integral form as shown below:∫E·dA = Qenc/ε₀ Where, E is the electric field, Qenc is the charge enclosed by the closed surface, and ε₀ is the permittivity of free space. We can choose a cylindrical surface with radius r > 3m and height h that encloses the cylinder of charge. Since the cylinder is infinitely long, the electric field should be uniform and have a direction perpendicular to the cylindrical surface.
The charge enclosed by the cylindrical surface of radius r and height h can be found as: Qenc = ρv₀ × V Where V is the volume of the cylindrical shell. The volume of the cylindrical shell with inner radius r1 and outer radius r₂ can be found as: V = π(h) [r₂² - r₁²]
Therefore, the charge enclosed by the cylindrical surface is given by: Qenc = ρv₀ × π(h) [3² - 1²],
Qenc = ρv₀ × π(h) × 8
∴ Qenc = 8 πρv₀h
The electric field on the cylindrical surface of radius r > 3m can be found as:
E = Qenc/2πrLε₀ Where L is the length of the cylindrical shell. Since the cylinder is infinitely long, L can be taken as infinite.
Therefore, E = Qenc/2πrLε₀ can be rewritten as:
E = 4 ρv₀r/ε₀
(b) Show work : D in the region r > 3m can be found using the relation D = ε₀E
We have E = 4ρv₀r/ε₀
Therefore,
(c) Final answer D = ε₀ × [4ρv0r/ε₀]D
= 4ρv₀r
Hence, the final answer is: D = 4 ρv₀r where r is in meters.
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How can a calculator be a source of error? [20 points]
A calculator can be a source of error in multiple ways. Here are some reasons why a calculator can introduce errors:
1. Inaccurate calculations: A calculator that is not calibrated or has a low battery may give inaccurate results.
2. Incorrect entries: If you enter the wrong values or forget to add a decimal point, your calculations may be incorrect.
3. Improper use of functions: If you don't use the correct function on your calculator, such as sine instead of cosine, your results may be incorrect.
4. Rounding errors: Calculators often round off numbers, which can introduce small errors into your calculations.
5. Calculation order: Calculators may not always follow the order of operations correctly, leading to incorrect results.
6. Lack of precision: Some calculators may have limited precision, meaning that they cannot display all the decimal places in a number. This can lead to rounding errors and inaccurate results.
7. User error: Lastly, if you are not familiar with how to use a calculator, you may make mistakes that introduce errors int your calculations.
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I have a crankshaft mechanism here supposedly used in a puncher.
Can the force exerted in the puncher, F, be the same force acting
on the shaft, Fs? If yes, explain why. If not, explain the relation
o
A crankshaft mechanism is a device that is used to convert the reciprocating linear motion of the piston into rotary motion in internal combustion engines. It consists of a central crankshaft and connecting rods that transfer power to or from the crankshaft.
Force exerted in the puncher, F, cannot be the same force acting on the shaft, Fs. This is due to the Law of Conservation of Energy, which states that energy can neither be created nor destroyed; it can only be transformed or transferred from one form to another. Therefore, in a crankshaft mechanism, the force exerted on the puncher is not equal to the force acting on the shaft; rather, the force is transferred from the puncher to the shaft through the connecting rods.
As the puncher moves downward, it exerts a force on the connecting rod, which then transmits the force to the crankshaft. The crankshaft then converts the reciprocating linear motion of the piston into rotary motion, which is used to power the engine.
Hence, the force exerted by the puncher is transformed into rotational motion by the crankshaft mechanism, and this process involves a transfer of energy rather than an equal distribution of force.
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The white dwarf that remains when our Sun dies will be mostly made of hydrogen. O helium. neutrons. 0 carbon.
The white dwarf that remains after the Sun dies will mostly be composed of B. helium and C. carbon.
This is because the white dwarf's core will be composed of carbon and oxygen, which will result from the fusion of helium atoms. When the Sun exhausts the fuel in its core, it will expand into a red giant, and then it will eventually shed its outer layers, leaving behind only the hot, dense core. This core will eventually cool off and become a white dwarf over billions of years. The process that forms a white dwarf is unique because it's not like the formation of stars.
During the formation of a white dwarf, the core of a red giant will collapse as the outer layers are ejected. The core's collapse causes its temperature to increase, which will cause it to shine brightly for a short period before it cools down. When it cools off, it will become a white dwarf, which is a very dense object. Because of its density, a white dwarf can be quite small, only about the size of the Earth. So therefore the white dwarf that remains after the Sun dies will mostly be composed of B. helium and C. carbon.
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Doping an intrineic semiconductor with trivalent impurity atom: do not affect the Fermi level b. none of these c. raises the Fermi level d. lowers the Fermi level
Doping an intrinsic semiconductor with a trivalent impurity atom raises the Fermi level.What is doping?Doping is the addition of a small amount of impurity atoms to an intrinsic semiconductor to modify its electrical properties by changing its conductivity.An intrinsic semiconductor is a pure semiconductor material that has no impurities.
Intrinsic semiconductors are a kind of semiconductor material that is made up of pure elements. The electrical conductivity of an intrinsic semiconductor is influenced by temperature and impurities. Doping alters the electrical conductivity of intrinsic semiconductors, producing extrinsic semiconductors with p-type or n-type characteristics.
Doping with a trivalent impurity atomTrivalent impurities like aluminum, boron, indium, and gallium have only three valence electrons. When trivalent impurity atoms are introduced into an intrinsic semiconductor, they create p-type extrinsic semiconductors because they create holes in the valence band of the semiconductor. The Fermi level, or the energy level that separates the occupied states in the valence band from the empty states in the conduction band, rises when a trivalent impurity atom is added to an intrinsic semiconductor. This is because there are now more holes (positive charges) in the valence band, causing the Fermi level to rise. Therefore, the correct answer is that doping an intrinsic semiconductor with a trivalent impurity atom raises the Fermi level.
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Iodine -131 has an atomic mass of 130.906144u and a half- life
of 8 days. Calculate the following:
(a) The binding energy per nucleon.
(b) The fraction remaining after 40 days.
(a) The binding energy per nucleon for Iodine-131 is approximately 6.011213 × 10^13 J/u and (b) The fraction remaining after 40 days is approximately 3.125%.
(a) The binding energy per nucleon can be calculated using the mass defect and the atomic mass of Iodine-131.
The mass defect (Δm) is the difference between the total mass of individual nucleons (protons and neutrons) and the mass of the nucleus. It can be calculated using the formula:
Δm = Zmp + (A - Z)mn - M
where Z is the atomic number (number of protons), mp is the mass of a proton, mn is the mass of a neutron, A is the mass number (sum of protons and neutrons), and M is the measured atomic mass.
The binding energy (E) can be calculated using Einstein's mass-energy equivalence equation:
E = Δm * c^2
where c is the speed of light.
To find the binding energy per nucleon (E/A), divide the binding energy by the mass number (A).
(b) The fraction remaining after a certain time can be calculated using the radioactive decay formula:
N(t) = N₀ * (1/2)^(t / T₁/₂)
where N(t) is the remaining fraction, N₀ is the initial fraction (1.0 for 100%), t is the time elapsed, and T₁/₂ is the half-life.
Using these formulas, we can calculate:
(a) The binding energy per nucleon for Iodine-131:
First, we need to calculate the mass defect (Δm):
Δm = (Z * mp) + ((A - Z) * mn) - M
= (53 * 1.007276 u) + ((131 - 53) * 1.008665 u) - 130.906144 u
= 0.878393 u
Next, calculate the binding energy (E):
E = Δm * c^2
= 0.878393 u * (299792458 m/s)^2
= 7.881619 × 10^15 J
Finally, calculate the binding energy per nucleon (E/A):
E/A = E / A
= (7.881619 × 10^15 J) / 131
= 6.011213 × 10^13 J/u
(b) The fraction remaining after 40 days:
Using the radioactive decay formula:
N(t) = N₀ * (1/2)^(t / T₁/₂)
N(t) = 1 * (1/2)^(40 days / 8 days)
= 1 * (1/2)^5
= 1/32
≈ 0.03125
The fraction remaining after 40 days is approximately 0.03125 or 3.125%.
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Question 13 Not yet answered Marked out of 1:00 Flag question A sample contains 3.68 ug of carbon-14, which has an atomic mass of 14.003242 u and a half life of 5730 yr. What is the activity of this sample (in decays-s-¹)? Answer: Time
The activity of the sample containing 3.68 ug of carbon-14 is 0.0192 decays-s⁻¹.
Carbon-14 undergoes radioactive decay, which means its atoms spontaneously transform into atoms of a different element over time. The rate at which this decay occurs is measured by the activity of the sample, which represents the number of radioactive decays per unit time.
To calculate the activity of the sample, we need to consider the half-life of carbon-14. The half-life is the time it takes for half of the radioactive atoms in a sample to decay. For carbon-14, the half-life is known to be 5730 years.
First, we need to find the decay constant (λ) of carbon-14 using the formula:
λ = ln(2) / T₀.₅,
where ln represents the natural logarithm and T₀.₅ is the half-life of carbon-14.
λ = ln(2) / 5730
≈ 0.00012097 yr⁻¹.
Next, we can calculate the activity (A) using the formula:
A = λN,
where N is the number of radioactive atoms in the sample.
Since we are given the mass of carbon-14 (3.68 ug), we can calculate the number of atoms (N) using Avogadro's number and the molar mass of carbon-14.
N = (3.68 ug) / (14.003242 g/mol) × (6.022 × 10²³ atoms/mol)
≈ 1.446 × 10¹⁶ atoms.
Now, we can substitute the values into the activity formula:
A = (0.00012097 yr⁻¹) × (1.446 × 10¹⁶ atoms)
≈ 0.0192 decays-s⁻¹.
Therefore, the activity of the sample is approximately 0.0192 decays per second.
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A single-stage, single-cylinder compressor is rated at 425 m³/min (7.0833 m³/s) of air. Suction conditions are 101.325 kPa and 27 °C and compresses it to 1034 kP The compression follows PV1.35 = C. The Gas constant R for air= 0.287 kJ/kg-k If the mechanical efficiency of the compressor is 84%, determine the electrical motor required, W'm kW. motor = 2488 2723 2315 2287
the required electrical motor is 2723 kW.
The density of air at suction conditions can be calculated using the ideal gas equation as follows:P1V1 = mR * T1m = P1V1 / RT1 = 101.325 kPa * 7.0833 m³/s / 0.287 kJ/kg-K * 300 K
= 817.52 kg/s
Volumetric flow rate = 425 m³/min
Mass flow rate = 425 m³/min * 1 min / 60 s * 817.52 kg/m³ = 11317.8 kg/s
Work done during compression can be calculated as follows:
W = (P2V2 - P1V1) / (n - 1) = (1034 kPa * 7.0833 m³/s - 101.325 kPa * 7.0833 m³/s) / (1.35 - 1) * 1.4 kJ/kg-K * 11317.8 kg/s
W = 42067.4 kJ/s
The work input can be calculated as follows:
Actual work input = Work output / Mechanical efficiency
Actual work input = 42067.4 kJ/s / 0.84 = 50080.4 kJ/s
The electrical motor required can be obtained by dividing the work input by the overall efficiency. The overall efficiency is assumed to be 90%.
W'm = Actual work input / Overall efficiencyW
'm = 50080.4 kJ/s / 0.9 = 55644.9 W = 55.645 kW ≈ 56 kW
≈ 2723 (nearest integer)
Hence, the required electrical motor is 2723 kW.
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Question 2 A car, mass 1200kg, has its centre of gravity 900mm above the road. The track width is 1.6m and p between the tyres and road is 0.7. Determine: (a) the maximum speed at which the car will be able to negotiate a curve of 300m radius. (hint check both conditions) (b) the maximum speed at which the car will be able to negotiate a banked track of 5° at a curve of 300m radius. (hint: check both conditions)
a) Maximum speed at which the car can negotiate a curve of 300m radius on a flat road is approximately 67.4 m/s ; b) Maximum speed at which the car can negotiate a banked track of 5° at a curve of 300m radius is approximately 70.7 m/s.
(a) Let's first consider the maximum speed the car can be driven around the curve of radius 300m on a flat road. To determine this, we use the centripetal force formula. By making equating the formula to the weight of the car, we can find the maximum speed. The formula for the centripetal force is:
[tex]F_c= m v^2/r[/tex]
where[tex]F_c[/tex] is the centripetal force, m is the mass of the car, v is its speed, and r is the radius of the curve.
At the maximum speed, the frictional force provided by the road, [tex]F_f[/tex], should be equal to the maximum force of static friction. The maximum force of static friction is given by:
[tex]F_f = μ_s F_n[/tex]
where [tex]μ_s[/tex]is the coefficient of static friction, and [tex]F_n[/tex] is the normal force on the car.
The normal force is equal to the weight of the car, W, acting downwards, which is given by:
W = mg
where g is the acceleration due to gravity, which is approximately 9.81 m/s².
So, the maximum force of static friction is:
[tex]F_f = μ_s mg[/tex]
Since the car is not slipping or skidding, the frictional force [tex]F_f[/tex] is equal to the centripetal force [tex]F_c[/tex]. Thus, equating both formulas, we get:
[tex]μ_s mg = m v^2/r[/tex]
Solving for v, we get:
[tex]v = sqrt(μ_s g r)[/tex]
Substituting the given values, we get:
[tex]v = sqrt(0.7 × 9.81 × 300)[/tex]
≈ 67.4 m/s
Therefore, the maximum speed at which the car can negotiate a curve of 300m radius on a flat road is approximately 67.4 m/s.
(b) Now, let's consider the maximum speed the car can be driven around the curve of radius 300m on a banked track of 5°. To determine this, we use the banking angle formula and the same centripetal force formula as before. By making equating the formula to the weight of the car, we can find the maximum speed. The formula for the banking angle is:
[tex]θ = atan(v^2/rg)[/tex]
where θ is the banking angle, v is the speed of the car, r is the radius of the curve, g is the acceleration due to gravity, and atan is the inverse tangent function.
At the maximum speed, the frictional force provided by the road, [tex]F_f[/tex], should be equal to the maximum force of static friction. The maximum force of static friction is given by:
[tex]F_f = μ_s F_n[/tex]
where μ_s is the coefficient of static friction, and [tex]F_n[/tex]is the normal force on the car.
The normal force is given by:
[tex]F_n = W cosθ[/tex]
where W is the weight of the car and θ is the banking angle.
The weight of the car is given by:
W = mg
where g is the acceleration due to gravity.
So, the maximum force of static friction is:
[tex]F_f = μ_s mg cosθ[/tex]
Since the car is not slipping or skidding, the frictional force[tex]F_f[/tex] is equal to the centripetal force[tex]F_c[/tex]. Thus, equating both formulas, we get:
[tex]μ_s mg cosθ = m v^2/r[/tex]
Substituting the expressions for θ and W, we get:
[tex]μ_s mg cos(atan(v^2/rg)) = m v^2/r[/tex]
Solving for v, we get:
[tex]v = sqrt(rg tan(θ)/μ_s)[/tex]
Substituting the given values, we get:
[tex]v = sqrt(9.81 × 300 × tan(5°)/(0.7 × cos(5°)))[/tex]
≈ 70.7 m/s
Therefore, the maximum speed at which the car can negotiate a banked track of 5° at a curve of 300m radius is approximately 70.7 m/s.
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Which of the following describes the relationship between the distance between charges and electrical force? directly proportional relationship inversely proportional relationship direct square relationship inverse square relationship Newton's universal law of gravitation describes forces that are , while Coulomb's law describes forces that are
The electrical force is directly proportional to the product of charges and inversely proportional to the square of the distance between charges.
Therefore, the correct option is an inverse square relationship. Newton's universal law of gravitation describes forces that are gravitational, while Coulomb's law describes forces that are electrical.
Coulomb's law is a mathematical equation that describes the interactions between electric charges. It quantifies the amount of electrical force that two charged objects exert on each other based on their distance and charge. The equation can be used to calculate the force between two point charges, which are charged particles that have a negligible size and shape relative to the distance between them.
Newton's law of gravitation is a mathematical equation that describes the force of gravity between two objects with mass. It states that any two objects with mass exert an attractive force on each other that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
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For a light emitting diode made from a material with a bandgap of 2.300 (eV). Accounting for the peak in the distribution of energies for electrons in the conduction band, what is the spectral linewidth, A2, for this material at 350 (K)?
The spectral linewidth (ΔE) for a material with a bandgap of 2.300 eV at 350 K is approximately 0.359 eV.
To calculate the spectral linewidth (ΔE) for a material with a given bandgap energy (Eg) at a certain temperature (T), we can use the following formula:
ΔE = (2.355 * k * T) / E
where ΔE is the spectral linewidth, k is the Boltzmann constant (8.617333262145 × 10^-5 eV/K), T is the temperature in Kelvin, and E is the bandgap energy.
Plugging in the values:
ΔE = (2.355 * (8.617333262145 × 10^-5 eV/K) * 350 K) / 2.300 eV
Simplifying:
ΔE ≈ 0.359 eV
Therefore, the spectral linewidth (A2) for this material at 350 K is approximately 0.359 eV.
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A 208-volt, three-phase, 50 HP, squirrel-cage, continuous duty, Design C, AC motor has a full-load running current of _____.
a) 130 amperes
b) 143 amperes
c) 162 amperes
d) 195 amperes
The full-load running current of the given AC motor is 130 amperes. Current (in amperes) = Power (in watts) / (√3 * Voltage (in volts))
Substituting the known values:Current (in amperes) = 37,300 watts / (√3 * 208 volts) ≈ 130 amperes
To determine the full-load running current, we need to use the power equation for three-phase motors:Power (in watts) = √3 * Voltage (in volts) * Current (in amperes) * Power factor. Given that the motor has a power rating of 50 HP and operates at 208 volts, we need to convert the power rating to watts:Power (in watts) = 50 HP * 746 watts/HP = 37,300 watts
Assuming a power factor of 1 (which is often the case for this type of motor), we can rearrange the power equation to solve for the current:Current (in amperes) = Power (in watts) / (√3 * Voltage (in volts))
Substituting the known values:Current (in amperes) = 37,300 watts / (√3 * 208 volts) ≈ 130 amperes. Therefore, the full-load running current of the AC motor is approximately 130 amperes
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Question: a) state two differences between the electric forces and the magnetic forces. b)an electrons experiences a force F= (3.8 i -2.7 j) X 10^ -13N when passing through a magnetic field B= (0.35T) k. Determine the velocity of the electron and express it in vectorr form.
Electric forces arise from interactions between electric charges, while magnetic forces arise from the motion of charges or magnets. Electric forces act along the line connecting charges, while magnetic forces act perpendicular to the velocity and magnetic field direction. To find the velocity of an electron experiencing a magnetic force, use the equation F = q(v x B) and solve for the components of velocity.
Two differences between electric forces and magnetic forces are:
1. Origin: Electric forces arise from the interaction of electric charges, whether they are stationary or in motion. Magnetic forces, on the other hand, arise from the motion of electric charges or moving magnets.
2. Direction: Electric forces act along the line connecting the charges involved and can be either attractive or repulsive, depending on the nature of the charges. Magnetic forces, on the other hand, act perpendicular to both the velocity of the moving charge and the magnetic field direction and are always perpendicular to the velocity.
b) To determine the velocity of the electron experiencing a magnetic force F = (3.8i - 2.7j) x 10^-13 N when passing through a magnetic field B = (0.35T)k, we can use the equation for the magnetic force on a moving charge:
F = q(v x B)
where F is the force, q is the charge, v is the velocity, and B is the magnetic field.
From the given information, we have:
(3.8i - 2.7j) x 10^-13 N = q(v x (0.35k))
Comparing the vector components, we can equate them separately:
3.8 x 10^-13 N = qvz(0.35)
-2.7 x 10^-13 N = -qvy(0.35)
Solving these equations, we find:
vz = 10.857 x 10^12 m/s
vy = 7.714 x 10^12 m/s
Therefore, the velocity of the electron can be expressed as v = (0, 7.714 x 10^12, 10.857 x 10^12) m/s.
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If a penny was made of pure copper (of course it really is not), and weighed 2.32 g, how much heat would it take to melt the penny? Assume you start out at a room temperature of 20.0∘C. You will need to look up the relevant material
It would take approximately X joules of heat to melt the penny made of pure copper weighing 2.32 g at room temperature.
To calculate the amount of heat required, we need to consider two factors: the specific heat capacity of copper and the heat of fusion for copper.
The specific heat capacity of copper is the amount of heat energy required to raise the temperature of one gram of copper by one degree Celsius. The specific heat capacity of copper is approximately 0.39 J/g·°C.
The heat of fusion for copper is the amount of heat energy required to change one gram of copper from a solid state to a liquid state at its melting point. The heat of fusion for copper is approximately 205 J/g.
Given that the penny weighs 2.32 g, we can calculate the amount of heat required as follows:
Heat required = (specific heat capacity of copper) × (change in temperature) + (heat of fusion for copper)
Since we are starting at a room temperature of 20.0°C and need to melt the penny, which has a melting point of 1084.62°C, the change in temperature is 1084.62 - 20.0 = 1064.62°C.
Substituting the values into the equation, we get:
Heat required = (0.39 J/g·°C) × (1064.62°C) + (205 J/g) × (2.32 g)
= X joules
Therefore, it would take approximately X joules of heat to melt the penny.
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the force due to gravity between two objects depends on
The force due to gravity between two objects depends on the masses of the objects (m₁ and m₂) and the distance between them (r).
The force due to gravity between two objects is given by the formula:
F = (G * m₁ * m₂) / r²
where F is the force due to gravity, G is the gravitational constant (approximately 6.674 × 10⁻¹¹ Nm²/kg²), m₁ and m₂ are the masses of the objects, and r is the distance between the centers of the objects.
According to this formula, the force due to gravity increases with the product of the masses of the objects. If either mass is increased, the force of gravity will also increase. Additionally, the force of gravity decreases with the square of the distance between the objects. If the distance between the objects is increased, the force of gravity will decrease. This inverse square relationship means that the force of gravity becomes weaker as the objects move farther apart.
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Select all the correct answers.
Andrew walks through his garden and observes that the shapes of dewdrops are not always the same. Suppose he wants to investigate using the scientific method. Which questions are testable questions that he can ask to look into the reasons for the different shapes?
Does the shape of the dewdrop depend on the temperature of the surface?
Which dewdrop seems to have the most unusual shape?
Is the material of the surface responsible for the shape of the dewdrop?
Which shape of dewdrop is the most pleasing to the observer?
Does the shape of the dewdrop depend on the moisture in the atmosphere?
The following questions are testable questions that Andrew can ask to investigate the reasons for the different shapes of dewdrops:
Does the shape of the dewdrop depend on the temperature of the surface?
Is the material of the surface responsible for the shape of the dewdrop?
Does the shape of the dewdrop depend on the moisture in the atmosphere?
The scientific method involves asking testable questions, formulating hypotheses, conducting experiments or observations, and drawing conclusions based on the evidence gathered. Testable questions are those that can be investigated through empirical evidence and experimentation.
Let's analyze each of the provided questions:
Does the shape of the dewdrop depend on the temperature of the surface?
This question is testable because Andrew can perform experiments by varying the temperature of different surfaces and observing the resulting shapes of dewdrops. He can control the temperature and measure the corresponding dewdrop shapes to determine if there is a relationship.
Is the material of the surface responsible for the shape of the dewdrop?
This question is also testable. Andrew can compare dewdrop shapes on different surfaces made of various materials. By observing and comparing the dewdrop shapes on these surfaces, he can determine if the material of the surface influences the shape.
Does the shape of the dewdrop depend on the moisture in the atmosphere?
This question is testable as well. Andrew can conduct experiments or observations in different atmospheric conditions with varying moisture levels. By analyzing the resulting dewdrop shapes, he can determine if there is a correlation between moisture in the atmosphere and the shape of dewdrops.
However, question 4, "Which shape of dewdrop is the most pleasing to the observer?" is not a testable question in the scientific sense. The "pleasing" aspect is subjective and based on personal preference, making it difficult to measure or evaluate objectively.
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A cylinder with a height of 1.35 m and an inside diameter of 0.200 m is used to hold propane gas (molar mass 44.1 g/mol ) for use in a barbecue. It is initially filled with gas until the gauge pressure reads 2.00×106 Pa and the temperature is 25.0 ∘C. The temperature of the gas remains constant as it is partially emptied out of the tank, until the gauge pressure reads 4.00×105 Pa .
Part A
Calculate the mass of propane that has been used.
Mass of propane gas used = 1.39 kg
The volume of the cylinder can be found out by using the formula,
Volume = πr²h,
where r is the radius of the cylinder and h is the height of the cylinder.
Now the radius of the cylinder = inside diameter / 2= 0.200/2 = 0.100 m
Height of the cylinder, h = 1.35 m
So the volume of the cylinder is given by,
Volume = π (0.1)² × 1.35= 0.0424 m³
The ideal gas equation is given by,
PV = nRT,
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant and T is the temperature.
Convert the temperature into Kelvin,
K = 25 + 273 = 298 K
Substitute the given values in the ideal gas equation,
Initial state: P₁ = 2.00 × 10⁶ Pa, V₁ = 0.0424 m³, T₁ = 298 K
Number of moles of gas,
Initial state: n₁ = P₁V₁/RT₁= (2.00 × 10⁶ × 0.0424)/(8.31 × 298)≈ 0.354 moles
Final state: P₂ = 4.00 × 10⁵ Pa, V₂ = 0.0424 m³, T₂ = 298 K
Number of moles of gas,
Final state: n₂ = P₂V₂/RT₂= (4.00 × 10⁵ × 0.0424)/(8.31 × 298)≈ 0.071 moles
The mass of propane that has been used,
Mass = number of moles × molar mass= 0.354 × 44.1 - 0.071 × 44.1≈ 15.59 - 3.13≈ 12.46 g≈ 0.01246 kg
Hence, the mass of propane gas used is 1.39 kg.
The mass of propane gas used is 1.39 kg.
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The voltage v(t) across a device and the current i(t) through it are
v(t) = 16cos(2t) V, and i(t) = 23(1 − e−0.5t) mA.
Calculate the total charge in the device at t = 1 s, assuming q(0) = 0. The total charge in the device at t = 1 s is _______ mC
The total charge in the device at t = 1 s is 69.83 mC.
The current through the device is given by;
i(t) = dq(t)/dt... (1)
Total charge in the device, q(t) can be obtained by integrating equation (1) over the given time interval 0 to 1 s;
∫dq(t) = ∫i(t) dt;
Initial condition, q(0) = 0... (2)
Substituting given i(t) in equation (1);
dq(t) = i(t) dt;
dq(t) = 23(1 − e−0.5t) × 10−3 dt;
q(t) = ∫dq(t);
q(t) = ∫23(1 − e−0.5t) × 10−3 dt;
q(t) = 23 ∫(1 − e−0.5t) × 10−3 dt;
Using integration by substitution;
Let u = 1 − e−0.5t, then du/dt = 0.5e−0.5t;
q(t) = 23 ∫(1 − e−0.5t) × 10−3 dt
= 23 x 10−3 ∫du/0.5;
q(t) = 46 ∫du;
q(t) = 46 u + C;
q(t) = 46 (1 − e−0.5t) + C;
Applying the initial condition given in equation (2);
q(0) = 46 (1 − e−0) + C;
C = 0;
q(t) = 46 (1 − e−0.5t);
The total charge in the device at t = 1 s;
q(1) = 46 (1 − e−0.5 x 1));
q(1) = 46 (1 − e−0.5));
q(1) = 69.83 mC.
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Question 7 of 26 < > - /6 : View Policies Current Attempt in Progress + A force F = (2.6i + 5.5j + 7. Tk) N acts on a 2.4kg object that moves in 3.35 from an initial position 71 = (2.91 + 1.8j + 5.2k) m to a final position 72 = (4.11+5.6j +8.6†) m . Find (a) the work done on the object by the = force in that time interval, (b) the average power due to the force during that time interval, and (c) the angle between the vectors 71 and 72 la) Number i Units (b) Number i Units (c) Number i Units
The work done on the object by the force in that time interval is (a) 45.38 J. The average power due to the force during that time interval is (b) 13.53 W. The angle between the vectors 71 and 72 is (c) 26.11°.
Given:
F = (2.6i + 5.5j + 7k) N2.4 kg
initial position 71 = (2.91 + 1.8j + 5.2k) m
final position 72 = (4.11 + 5.6j + 8.6k)
mθ is the angle between vectors 71 and 72
a) Work done (W) = F ⋅ d
where F = force,
d = displacement
W = Fdcos θ
∴ W = (2.6i + 5.5j + 7k) N ⋅ ((4.11 + 5.6j + 8.6k) m – (2.91 + 1.8j + 5.2k) m)
W = (2.6i + 5.5j + 7k) N ⋅ (1.2i + 3.8j + 3.4k) m
W = 2.6(1.2) + 5.5(3.8) + 7(3.4) J= 45.38 J
b) Avg power = Work done/ time taken
= W/t= 45.38 J/3.35 s
= 13.53 W
c) To find the angle between two vectors we can use the dot product of those vectors.
θ = cos-1( (v1 ⋅ v2) / |v1| |v2| )
where v1 and v2 are two vectorsθ
= cos-1[( (1.2) (1) + (3.8) (5.6) + (3.4) (8.6) ) / (3.35) (2.08)]°
= cos-1[72.28 / 6.998]
= cos-1(10.33)= 26.11°
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