Answer:
P1 = 1.3 (500 + 60) = 728 kg-m total momentum to right at start
P2 = (v2 - 10) 60 + 500 v2
total momentum after running at -10 with respect to cart = 728 where v2 is the new speed of the cart
728 = 560 v2 - 600
v2 = 1328 / 560 = 2.37 m/s new speed of cart
Check:
After: p2 for cart = 500 * 2.37 = 1186
p1 for man = (2.37 - 10) * 60 = -458
P2 = p1 + p2 = 728 total momentum unchanged
Two ends of a steel wire of length 8m and 2mm radius are fixed to two rigid supports. Calculate the increase in tension when the temperature falls by 10°C. Given linear expansivity of steels = 12x10^_6 per kelvin and Young's modules for steel =2x10^11 n/m^2
The increase in tension on the steel wire is 8,484.75 N.
The given parameters;
original length of the wire, l = 8 mradius of the wire, r = 2 mmThe area of the steel wire is calculated as follows;
[tex]A = \pi r^2\\\\A = \pi \times (2\times 10^{-3})^2\\\\A = 1.257 \times 10^{-5} \ m^2[/tex]
The extension of the steel wire is calculated as follows;
[tex]\Delta l = \alpha \times l\times \Delta T\\\\\Delta l = (12\times 10^{-6}) \times (8) \times (10 + 273)\\\\\Delta l = 0.027 \ m[/tex]
The increase in tension on the steel wire is calculated as follows;
[tex]E = \frac{stress}{strain } = \frac{\ F/A}{\Delta l/l} \\\\E = \frac{F\times l}{A \times \Delta l} \\\\F = \frac{E\times A \times \Delta l }{l} \\\\F = \frac{(2\times 10^{11}) \times (1.257\times 10^{-5})\times 0.027}{8} \\\\F = 8,484.75 \ N[/tex]
Thus, the increase in tension on the steel wire is 8,484.75 N.
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Which light is most sensitive to the eyes?
Answer:
Our eyes are most sensitive to the wavelengths corresponding to the yellow and green colors of the spectrum. Flashy signs and some fire engines are painted in a yellowish-green color to attract our attention.
A 30.0-g object moving to the right at 20.5 cm/s overtakes and collides elastically with a 13.0-g object moving in the same direction at 15.0 cm/s. Find the velocity of each object after the collision. (Take the positive direction to be to the right. Indicate the direction with the sign of your answer.)
Let m₁ and m₂ be the masses of the two objects, and v₁ and v₂ their initial velocities. So
m₁ = 30.0 g = 0.0300 kg
m₂ = 13.0 g = 0.0130 kg
v₁ = + 20.5 cm/s = 0.205 m/s
v₂ = + 15.0 cm/s = 0.150 m/s
and we want to find v₁' and v₂', the final velocities of either object after their collision.
Momentum is conserved throughout the objects' collision, so that
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where v₁' and v₂' are the first and second object's velocities after the collision.
Kinetic energy is also conserved, so that
1/2 m₁v₁² + 1/2 m₂v₂² = 1/2 m₁(v₁')² + 1/2 m₂(v₂')²
or
m₁v₁² + m₂v₂² = m₁(v₁')² + m₂(v₂')²
From the first equation (omitting units), we have
0.0300 • 0.205 + 0.0130 • 0.150 = 0.0300 v₁' + 0.0130 v₂'
0.0810 = 0.0300 v₁' + 0.0130 v₂'
81 = 30 v₁' + 13 v₂'
From the second equation,
0.0300 • 0.205² + 0.0130 • 0.150² = 0.0300 (v₁')² + 0.0130 (v₂')²
0.00155 ≈ 0.0300 (v₁')² + 0.0130 (v₂')²
1.55 ≈ 30 (v₁')² + 13 (v₂')²
Solving both equations simultaneously gives two solutions, one of which corresponds to the initial conditions. The other yields
v₁' ≈ + 0.172 m/s
and
v₂' ≈ + 0.227 m/s
when a torque is acting on a fly wheel the angular velocity of the fly wheel changes from 10rad/sec to 25rad/sec in 5sec.what will be the magnitudes of the angular acceleration of the fly wheel?
Hello!
We can use the following angular kinematic equation to solve:
α = Δω/Δt, or (ωf-ωi)/t
Plug in the given values:
α = (25 - 10)/5 = 15/5 = 3 rad/sec²
You have a 25 W and a 100 W bulb at home. When you connect one or the other, which bulb carries the greater current
Answer:
100 w bulb has more current.
Explanation:
P=V^2/R.
when velocity is constant power is inversly proportional to resistence, so 25 W will have an hogher resistance tjan a 100 w bulb.
P=VI
when v is constant, power is directly proportional to I. hence, 100 w bulbs will carry more current.
When you connect one or the other, the bulb that's connected carries more current than the one that's not connected.
When you connect BOTH of them, the 100W bulb carries more current than the 25W one.
A plastic bag is massed. It is then filled with a gas which is insoluble in water and massed again. The apparent weight of the gas is the difference between these two masses. The gas is squeezed out of the bag to determine its volume by the displacement of water. What is the actual weight of the gas
The actual weight of the gas = apparent weight + weight.
The actual weight = [tex]W_{A}[/tex] + W
Given that a plastic bag is massed. It is then filled with a gas which is insoluble in water and massed again.
If the apparent weight of the gas is the difference between these two masses, then let the apparent weight = [tex]W_{A}[/tex]
The gas is squeezed out of the bag to determine its volume by the displacement of water. Since
density = mass / volume
The density of water is 1000 kg/[tex]m^{2}[/tex]
we can get the mass of the gas by making m the subject of the formula.
W = mg
The actual weight of the gas = apparent weight + weight
That is,
The actual weight = [tex]W_{A}[/tex] + W
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Two parallel plates, separated by 0.20 m, are connected to a 12-V battery. An electron released from rest at a location 0.10 m from the negative plate. When the electron arrives at a distance 0.050 m from the positive plate, what is the potential difference between the initial and final points
The potential difference between the initial and final point is 3.0 V.
The given parameters:
distance between the plates, d = 0.2 mvoltage across the plates, V = 12 Vposition of the electron from negative plate, x₁ = 0.1 mposition of the electron from the positive plate, x₂ = 0.0 5mThe potential difference between the initial and final point is calculated as follows;
[tex]E = \frac{V}{d} \\\\\frac{V_1}{d_1} = \frac{V_2}{d_2}[/tex]
where:
[tex]d_2[/tex] is the distance of the electron between the positive and negative plate
[tex]0.1 + d_2 + 0.05 = 0.2\\\\d_2 + 0.15 = 0.2\\\\d_2 = 0.2 - 0.15\\\\d_2 = 0.05 \ m[/tex]
[tex]V_2 = \frac{V_1d_2}{d_1} \\\\V_2 = \frac{12 \times 0.05}{0.2} \\\\V_2 = 3.0 \ V[/tex]
Thus, the potential difference between the initial and final point is 3.0 V.
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What was different about the molecules you needed to make protein 3 compared to the molecules you used to make protein 2?
Answer:
the different about the molecules we needed to make protein 3 compared to the molecules we used to make protein 2 is that if we used 2 molecules than it will be smaller than using protein 3.
A. A small, closed chamber of gas is heated. When the gas in the
chamber expands, it does 5 J of work on a piston. The gas has
an initial energy of 8 J and a final energy of 30 J. Considering
the equation for the first law of thermodynamics (AU = Q + W.
does the work done in this scenario have a positive or negative
value? Explain.
B. How much heat is added to the gas in the chamber?
A.
The work has a negative value.
Since the gas in the chamber expands, it increases in volume and does positive work on the piston since the change in volume is positive.
Work, W = pΔV. Since the gas expands, ΔV > 0. So, W > 0. Thus Work done by the gas is positive.
Since in ΔU = Q + W, W here is work done by the surroundings, W is negative since it is the opposite of the work done on the surroundings by the gas..
So, the work has a negative value.
B.
The heat added to the gas chamber is 17 J
From the first law of thermodynamics, ΔU = Q + W where ΔU = internal energy change = U₂ - U₁ where U₁ = initial energy = 8 J and U₂ = final energy = 30 J, Q = heat added to the gas chamber and W = work on piston = - 5 J
ΔU = Q + W
U₂ - U₁ = Q + W
30 J - 8 J = Q + (-5 J)
22 J = Q - 5 J
Q = 22 J + 5 J
Q = 27 J
The heat added to the gas chamber is 27 J.
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PLEASE HELP BRAINLIEST TO THE RIGHT ANSWER!!!!!
A 7.8 kg object is suspended by a string from the ceiling of an elevator. The acceleration of gravity is 9.8 m/s^2.
A. Determine the tension in the string if it is accelerating upward at a rate of 1.5 m/s^2. Answer in units of N.
B. Determine the tension in the string if it is accelerating downward at a rate of
1.5 m/s^2
Gravitational pull downward on the object: 7.8 x 9.8 = 76.44N
A. 7.8 x 1.5 = 11.7N upward force
Tension = 76.44 + 11.7 = 88.14 N
Answer: 88.14 N
B. 76.44 - 11.7 = 64.74N
Answer: 64.74 N
Answer:
88.14 N
64.74 N
Explanation:
When measuring the critical angle, in which medium do we need the refracted ray to be (air or glass)
Answer:
N1 sin theta1 = N2 sin theta2 Snell's Law
For the refracted ray to be reflected then sin theta2 = 1
N1 sin theta 1 = N2
Also N2 must be less than N1 for complete reflection
sin theta1 = N2/N1
If you are considering air and glass then N2 = 1 (for air)
sin theta1 = 1 / N2 where N2 must be for glass in this case
7.2
8
How is kinetic
molecular
model of matter helpful in
differentiating various states
of matter?
V
Answer:
The kinetic molecular theory of matter asserts that: Matter is made up of particles that are continually moving. All particles contain energy, however the energy fluctuates depending on the temperature the sample of matter is in. This in turn decides whether the material exists in the solid, liquid, or gaseous form.
Explanation:
Hope it helps:)
You throw an 8.75 g coin straight up into the air. The coin travels a distance of 337 cm upward. What was the initial speed of the coin when you released it
Answer:
hi
Explanation:
i like saying hi
A double-pane glass window is 60.0 cm x 90.0 cm and has 3.00-mm window panes. If the temperature difference between inside and outside is 24.0 K, how far apart should the panes be to have a heat loss of 4.09 W? Assume there is air in the gap.
The distance between the glass to have the given heat loss is 2.54 m.
The given parameters:
dimension of the window, = 60 cm by 90 cmtemperature, T = 24 Kheat lost, Q = 4.09 Wthermal conductivity of glass, k = 0.8 W/mKThe area of the glass window is calculated as follows;
[tex]A = 0.6 \times 0.9\\\\A = 0.54 \ m^2[/tex]
The distance between the glass is calculated as follows;
[tex]Q = \frac{KA \Delta T}{\Delta x} \\\\\Delta x = \frac{kA \Delta T}{Q} \\\\\Delta x = \frac{0.8 \times 0.54 \times 24 }{4.09} \\\\\Delta x = 2.54 \ m[/tex]
Thus, the distance between the glass to have the given heat loss is 2.54 m.
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What is Rest in physics ?
Rest in physics generally refers to the state in which an object is stationary.
Rest in physics refers to a situation in which an object does not move from one point to another. Usually, an object is at rest when it is acted upon by equal and opposite forces. For example, a book lying on a table.In a nutshell, the state of rest in physics generally refers to the state in which an object is stationary and does not translate.Hence, in Physics, an object is in a state of rest when it does not move from one point to another.
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Three point charges are placed on the y-axis: a chargeqaty=a,a charge-2qat the origin, and a chargeqaty= -a.Such an arrangement is called an electricquadrupole.(a) Find the magnitude and direction of the electric field at points on thepositive x-axis. (b) Use the binomial expansion to find an approximate expression for the electric field valid for x >> a.Contrast this behavior to that of the electric field of a point charge and that of the electric field of a dipole.
which of the following statements describes a perfectly inelastic collision
Answer: An ice hockey player picks up a trophy as he slides past it.
Two pool balls colliding precisely inelastically and rebounding off one another is an example. So, the correct option is A.
What is an Inelastic collision?Inelastic collisions are those in which the total kinetic energy is lower after the impact than it was before. The stick is travelling quickly in the direction of the ball before to the contact. The stick comes to a stop following the accident. It transfers some of its kinetic energy to the cue ball, which rolls forward.
The type of collision mentioned in the given example is known as an inelastic collision is one in which the system's momentum is preserved but its kinetic energy is not. In a ballistic pendulum, an example of an inelastic collision. Dropped ball of clay is another illustration of an inelastic collision.
Thus, the correct option is A.
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Your question is incomplete, most probably the complete question is:
Which of the following statements describes a perfectly inelastic collision?
A. Two billiard balls bounce off of each other,
B. A car crashes into a tree and rebounds in the other direction,
C. A wad of chewing gum is thrown and sticks to a moving truck,
D. Two ice skaters hit each other and fall over in opposite directions,
 A wooden box with a mass of 10.0 kg rest on a ramp that is incline at an angle of 25° to the horizontal. A rope attached to the box runs parallel to the ramp and then passes over a frictionless bully. A bucket with a mass of M hangs at the end of the rope. The coefficient of static friction between the ramp in the box is 0.50. The coefficient of Connecticut friction between the ramp in the box is 0.35.
Suppose the box remains at rest relative to the ramp. What is the maximum magnitude of the friction force exerted on the box by the ramp?
The maximum magnitude of the friction force exerted on the box by the ramp is 44.41 N.
The given parameters;
Mass of the box, m = 10 kgInclination of the ramp, θ = 25⁰Coefficient of static friction, μ = 0.5 Coefficient of kinetic friction, μk = 0.35The normal force on the wooden box is calculated as follows;
[tex]F_n = mg \times cos(\theta)\\\\F_n = 10 \times 9.8 \times cos(25)\\\\F_n = 88.8 2 \ N[/tex]
The maximum magnitude of the friction force exerted on the box by the ramp is calculated as follows;
[tex]F_f = \mu \times F_n\\\\F_f = 0.5 \times 88.82 \\\\F_f = 44.41 \ N[/tex]
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Your in an escape room. Only its real and your oxygen runs out in 3 minutes unless you solve this problem, The problem :a ball is thrown and travels 30 inches before bouncing, It bounces and travels 50% of the distance traveled. It continues to do this until coming to rest. What is the total distance traveled by ball? Also it's cold and dark and you have no phone nor anyone to ask. You have no ball
60 inches
because if 30=50% then 60=100% so its 60 inches
The sum of the series allows to find the result for the total distance that the ball bounces is:
total distance = 59.52 in
A series is a set of things or numbers related by a specific operation.
They indicate that the ball falls from an initial height y₀ = 30 in. and it bounces 50% of the height and the process is repeated until it stops, see attached.
Let's build a table to observe the sequence.
drop height rebound
1 30 15
2 15 7.5
3 7.5 3.75
If we call the first term y₀
The first bounce can be found.
y₁ = [tex]\frac{y_o}{2}[/tex]
The second bounces.
[tex]y_2 = \frac{y_1}{2} \\y_2 = \frac{y_o}{4}[/tex]
The third bounce.
[tex]y_3 = \frac{y_2}{2} \\y_3 = \frac{y_0}{ 8}[/tex]
By observing this table we can construct a series of the form
Total distance = [tex]y_o \ ( 1 + \frac{1}{2} + \frac{1}{4}+ \frac{1}{8} + ... +\frac{1}{2n} )[/tex]
The sum of the serie has a result of
sum = 127/64 = 1,984
Let's calculate
distance total = 30 1,984
Distance total = 59.52 cm
In conclusion, using the sum of the series we can find the result for the total distance that the ball bounces is:
total distance = 59.52 in
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A child at the top of a slide de has a gravitational store of 1800J. What is the child's maximum kinetic store as he slides down? Explain why
Hi there!
We know that:
Initial Total Mechanical Energy = Final Total Mechanical Energy
(Ei = Ef)
In this instance:
Ei = Gravitational Potential Energy
Ef = Kinetic Energy
In the absence of friction, ALL of the initial potential energy will be changed into kinetic energy at the bottom of the slide. Thus, the maximum kinetic energy of the child will be 1800 J.
5) You pull a 10.0 kg wagon along a flat road. You exert a force of 80.0 N at an angle of 30.0 degrees above the horizontal while you move the wagon 10.0 m forward. The coefficient of friction between the wagon and road is 0.500. Calculate the work down by you and the work done by friction.
Consult the attached free body diagram. The only forces doing work on the wagon are the frictional force opposing the wagon's motion and the horizontal component of the applied force.
By Newton's second law, the net vertical force is
• ∑ F [v] = n + (80.0 N) sin(30.0°) - mg = 0
where a is the acceleration of the wagon.
Solve for n (the magnitude of the normal force) :
n = (10.0 kg) g - (80.0 N) sin(30.0°) = 58.0 N
Then
f = 0.500 (58.0 N) = 29.0 N
Meanwhile, the horizontal component of the applied force has magnitude
(80.0 N) cos(30.0°) ≈ 69.3 N
Now calculate the work done by either force.
• friction: -(29.0 N) (10.0 m) = -290. J
• pull: (69.3 N) (10.0 m) = 693 J
Be able to list the three compounds that are formed as products of highly exothermic
reactions such as detonating nitrogen-based explosives?
Answer:
Ammonium perchlorate NH4ClO4
Ammonium Nitrate
Calcium Cyanamide
When detonated, the reaction products are all gases, such as water vapor, nitrogen gas, and oxides of nitrogen.
Hopefully this helps :)
1. It’s fall and time for the corn maze and bonfire and you just can’t wait. On your way to the farm though a turkey flies out in front of you, so you slam on the brakes and go from from 30.0 m/s to 18.0 m/s. Luckily your date brought a stop watch and told you the whole thing took place in 10.5s. What is your acceleration and how far did you go?
Acceleration = (change in velocity ( final speed - starting speed))/ (time)
Acceleration = (18-30)/10.5
Acceleration = -12/10.5
Acceleration = -1.14 m/s^2
Distance = 30m/s x 10.5s + 1/2(1.14)(10.5)^2
Distance = 252.2 meters
Heat is transferred from molecules with more kinetic energy to molecules with _________kinetic energy.
Answer:
less
Explanation:
heat travels from hotter to colder object. hotter object has more kinetic energy.
We have seen in earlier readings how to determine the speed of a wave on a string. What will happen to the wavelength of a sinusoidal wave on a string if the tension in the string is increased (assuming we keep the frequency of the wave the same)
This question involves the concepts of tension in a string and the wavelength of the wave in a string.
The wavelength of a sinusoidal wave will "increase by square power" on a string if the tension in the string is increased when the frequency is kept constant.
The speed of a wave on a string is given by the following formula:
[tex]v=\sqrt{\frac{T}{\mu}}[/tex]
where,
v = speed of wave = fλ
f = frequency of the wave
λ = wavelength of the wave
T = tension force
μ = linear mass density of the string
Therefore,
[tex]f\lambda=\sqrt{\frac{T}{\mu}}\\\\T = f^2\lambda^2\mu[/tex]
It is given that the frequency is kept constant. The linear mass density is also constant for a string. Therefore,
[tex]T=(constant)\lambda^2\\T\ \alpha\ \lambda^2[/tex]
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The Air Force is F 22 raptor fighter jets mass is 21,000 KG the F 22 is flying at a height of 26,000 miles what is its gravitational potential energy
Answer:
5460000000 J OR 5460000 KJ
Explanation:
GPE = mgh
21000*10*26000
=5460000000J OR 5460000KJ
A Projectile of mass 0.277 kf is shot from a cannon. The end of the cannon's barrel is at heiht 6.8 m. The initial velocity of the projectile has a horizontal component of 6.1 m/s. The projectile rises to a maximum height above the end of the cannon's barrel and strikes th ground a horizontal distance past the end of the cannon;s barrel. What is the time it takes for the projectile to reach its maximm height? The acceleration of gravity is 9.8 m/s^2.
The time taken by the projectile to reach its maximum height is 0.62 second.
Given parameters:
Horizontal component of initial velocity of the projectile: [tex]u_x[/tex] = 6.1 m/s.
To reach maximum height, the projectile should be throwed in 45 degree. So, vertical component of initial velocity of the projectile: [tex]u_y[/tex] = 6.1 m/s.
Given, The acceleration of gravity is 9.8 m/s².
At maximum height the vertical component of velocity of the projectile becomes zero due to acceleration due to gravity acts downwards.
So, from formula [tex]v_y[/tex] = [tex]u_y[/tex] - gt; we can write:
0 = 6.1 - 9.8t
⇒ t = 6.1/9.8 = 0.62 s.
Hence, the time taken by the projectile to reach its maximum height is 0.62 second.
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An electron is held up against the force of gravity by the attraction of a fixed proton some distance above it. How far above the electron is the proton
5.08 m
Explanation:
The weight of the electron is being counteracted by the attractive electrostatic force exerted by the proton above it. We can write the force equation as follows:
[tex]m_eg = k_e\dfrac{Q_pQ_e}{r^2}[/tex]
where the Q's are the charges of the proton and electron, r is the distance between the particles, g is the acceleration due to gravity, [tex]m_e[/tex] is the mass of the electrons and [tex]k_e[/tex] is the Coulomb constant. So solving for r, we get
[tex]r^2 = k_e\dfrac{Q_pQ_e}{m_eg}[/tex]
Taking the square root of r^2, we then get the distance as
[tex]r = \sqrt{k_e\dfrac{Q_pQ_e}{m_eg}}[/tex]
The values are given as follows:
[tex]m_e = 9.11×10^{-31}\:\text{kg}[/tex]
[tex]g = 9.8\:\text{m/s}^2[/tex]
[tex]Q_p = Q_e = 1.60×10^{-19}\:\text{C}[/tex]
[tex]k_e = 8.99×10^9\:\text{N-m}^2\text{/C}^2[/tex]
Putting in all of these values in our equation for r,
[tex]r = \sqrt{\dfrac{(8.99×10^9\:\text{N-m}^2\text{/C}^2)(1.60×10^{-19}\:\text{C})^2}{(9.11×10^{-31}\:\text{kg})(9.8\:\text{m/s}^2)}}[/tex]
[tex]\:\:\:\:\:= 5.08\:\text{m}[/tex]
A glass beaker of unknown mass contains of water. The system absorbs of heat and the temperature rises as a result. What is the mass of the beaker? The specific heat of glass is 0.18 cal/g ∙ °C, and that of water is 1.0 cal/g ∙ C°.
From the information provided in the question, the mass of the beaker is 144.4 g.
From the information provided in the complete question;
volume of water = 74 mL
Mass of water = 74 g
specific heat of glass = 0.18 cal/g ∙ °C
specific heat of water = 1.0 cal/g ∙ C°
Mass of glass = x g
Total heat gained by the system = 2000.0cal
Temperature rise = 20.0°C
Heat gained by system = Heat gained by glass + Heat gained by water
Heat gained by glass = x × 0.18 × 20
Heat gained by water = 74 × 1.0 × 20
Hence;
2000 = (x × 0.18 × 20) + ( 74 × 1.0 × 20)
2000 - 1480 = (x × 0.18 × 20)
x = 520/3.6
x = 144.4 g
Missing parts;
A glass beaker of unknown mass contains 74.0 ml of water. The system absorbs 2000.0cal of heat and the temperature rises 20.0°C as a result. What is the mass of the beaker? The specific heat of glass is 0.18
cal/g °C, and that of water is 1.0 cal/g °C.
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Eric drops a 2.20 kg water balloon that falls a distance of 45.08 m off the top of a
building. What is the kinetic energy at the bottom?
Answer:
972 J
Explanation:
At the bottom, all the gravitational potential energy was converted into kinetic energy. If you calculate the GPE, its value will be the same that the KE at the bottom. The GPE can be calculated this way:
GPE = mass×gravity×heigth
GPE = 2.2×9.8×45.08 ≈ 972
