Answer:
Magnetic Forces on Moving Charges. The magnetic force on a free moving charge is perpendicular to both the velocity of the charge and the magnetic field with direction given by the right hand rule.
The direction of the charge where it does not experience magnetic force is left and right. The correct option is C and D.
What is a magnetic field?A magnetic field is a region in space where a magnetic force can be observed. It is created by moving electric charges, such as electrons, and is characterized by the direction and strength of the force it exerts on other magnetic materials or moving charges.
Magnetic field lines are used to visualize the direction and strength of the magnetic field. They represent the path that a small magnetic north pole would follow if placed in the magnetic field. The direction of the magnetic field is given by the direction in which the north pole of a compass needle would point if placed in the field.
Magnetic flux is the measure of the strength of the magnetic field passing through a surface. It is given by the product of the magnetic field strength and the area of the surface, as well as the cosine of the angle between the magnetic field and the surface normal. The unit of magnetic flux is Weber (Wb).
Magnetic flux is important in many applications, such as electric motors and generators, where the interaction between the magnetic field and moving charges produces electrical energy. It is also used in magnetic imaging techniques, such as MRI, to visualize the internal structures of the human body.
Here in the question,
Options A (up), B (down), E (into the screen), and F (out of the screen) are all perpendicular to the direction of the magnetic field and so will experience a magnetic force.
Therefore, options C and D are correct i. e left and right.as they are parallel and anti-parallel to the direction of the magnetic field, respectively.
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Determine the density in kg \cm of solid whose Made is 1080 and whose dimension in cm are length=3 ,width=4,and height=3
Answer:
d = 30kg/cm³
Explanation:
d = m/v
d = 1080kg/(3cm*4cm*3cm)
d = 30kg/cm³
When using the lens equation, a negative value as the solution for di indicates that the image is
Answer:
The Anatomy of a Lens
Refraction by Lenses
Image Formation Revisited
Converging Lenses - Ray Diagrams
Converging Lenses - Object-Image Relations
Diverging Lenses - Ray Diagrams
Diverging Lenses - Object-Image Relations
The Mathematics of Lenses
Ray diagrams can be used to determine the image location, size, orientation and type of image formed of objects when placed at a given location in front of a lens. The use of these diagrams was demonstrated earlier in Lesson 5 for both converging and diverging lenses. Ray diagrams provide useful information about object-image relationships, yet fail to provide the information in a quantitative form. While a ray diagram may help one determine the approximate location and size of the image, it will not provide numerical information about image distance and image size. To obtain this type of numerical information, it is necessary to use the Lens Equation and the Magnification Equation. The lens equation expresses the quantitative relationship between the object distance (do), the image distance (di), and the focal length (f)
A 90 kg man stands in a very strong wind moving at 17 m/s at torso height. As you know, he will need to lean in to the wind, and we can model the situation to see why. Assume that the man has a mass of 90 kg, with a center of gravity 1.0 m above the ground. The action of the wind on his torso, which we approximate as a cylinder 50 cm wide and 90 cm long centered 1.2 m above the ground, produces a force that tries to tip him over backward. To keep from falling over, he must lean forward.
A. What is the magnitude of the torque provided by the wind force? Take the pivot point at his feet. Assume that he is standing vertically. Assume that the air is at standard temperature and pressure.
B. At what angle to the vertical must the man lean to provide a gravitational torque that is equal to this torque due to the wind force?
Answer:
a) [tex]t=195.948N.m[/tex]
b) [tex]\phi=13.6 \textdegree[/tex]
Explanation:
From the question we are told that:
Density [tex]\rho=1.225kg/m^2[/tex]
Velocity of wind [tex]v=14m/s[/tex]
Dimension of rectangle:50 cm wide and 90 cm
Drag coefficient [tex]\mu=2.05[/tex]
a)
Generally the equation for Force is mathematically given by
[tex]F=\frac{1}{2}\muA\rhov^2[/tex]
[tex]F=\frac{1}{2}2.05(50*90*\frac{1}{10000})*1.225*17^2[/tex]
[tex]F=163.29[/tex]
Therefore Torque
[tex]t=F*r*sin\theta[/tex]
[tex]t=163.29*1.2*sin90[/tex]
[tex]t=195.948N.m[/tex]
b)
Generally the equation for torque due to weight is mathematically given by
[tex]t=d*Mg*sin90[/tex]
Where
[tex]d=sin \phi[/tex]
Therefore
[tex]t=sin \phi*Mg*sin90[/tex]
[tex]195.948=833sin \phi[/tex]
[tex]\phi=sin^{-1}\frac{195.948}{833}[/tex]
[tex]\phi=13.6 \textdegree[/tex]
At room temperature, sound travels at a speed of about 344 m/s in air. You see a distant flash of lightning and hear the thunder arrive 7.5 seconds later. How many miles away was the lighting strike? (Assume the light takes essentially no time to reach you).
Answer:
1.6031 miles
Explanation:
Given the following data;
Speed = 344 m/s
Time = 7.5 seconds
To find how many miles away was the lighting strike;
Mathematically, the distance travelled by an object is calculated by using the formula;
Distance = speed * time
Distance = 344 * 7.5
Distance = 2580 meters
Next, we would have to convert the value of the distance travelled in meters to miles;
Conversion:
1609.344 metres = 1 mile
2580 meters = X mile
Cross-multiplying, we have;
X * 1609.344 = 2580
X = 2580/1609.344
X = 1.6031 miles
The New England Merchants Bank Building in Boston is 152 m high. On windy days it sways with a frequency of 0.18 Hz , and the acceleration of the top of the building can reach 1.9 % of the free-fall acceleration, enough to cause discomfort for occupants.
Required:
What is the total distance, side to side, that the top of the building moves during such an oscillation?
Answer:
The total distance, side to side, that the top of the building moves during such an oscillation is approximately 0.291 meters.
Explanation:
Let suppose that the building is experimenting a Simple Harmonic Motion due to the action of wind. First, we determine the angular frequency of the system ([tex]\omega[/tex]), in radians per second:
[tex]\omega = 2\pi\cdot f[/tex] (1)
Where [tex]f[/tex] is the frequency, in hertz.
If we know that [tex]f = 0.18\,hz[/tex], then the angular frequency of the system is:
[tex]\omega = 2\pi\cdot (0.18\,hz)[/tex]
[tex]\omega \approx 1.131\,\frac{rad}{s}[/tex]
The maximum acceleration experimented by the system is represented by the following formula, of which we estimate amplitude of the oscillation:
[tex]r\cdot g = \omega^{2}\cdot A[/tex] (2)
Where:
[tex]r[/tex] - Ratio of real acceleration to free-fall acceleration, no unit.
[tex]g[/tex] - Free-fall acceleration, in meters per square second.
[tex]A[/tex] - Amplitude, in meters.
If we know that [tex]\omega \approx 1.131\,\frac{rad}{s}[/tex], [tex]r = 0.019[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the amplitude of the oscillation is:
[tex]A = \frac{r\cdot g}{\omega^{2}}[/tex]
[tex]A = \frac{(0.019)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{\left(1.131\,\frac{rad}{s} \right)^{2}}[/tex]
[tex]A \approx 0.146\,m[/tex]
The total distance, side to side, is twice the amplitude, that is to say, a value of approximately 0.291 meters.
calculate the force on an object with mass of 50kg and gravity of 10
A cylindrical wire made of an unknown alloy hangs from a support in the ceiling. You measure the relaxed length of the wire to be 16 m long; and the radius of the wire to be 3.5 m. When hang a 5 kg mass from the wire, you measure that it stretches a distance of 4 x 10 m The average bond length between atoms is 2.3 x 10^0 m for th alloy.
Required:
What is the stiffness of a typical interatomic bond in the alloy
Answer: hello some of your values are wrongly written hence I will resolve your question using the right values
answer:
stiffness = 1.09 * 10^-6 N/m
Explanation:
Given data:
Length ( l ) = 16 m
radius of wire ( r ) = 3.5 m
mass ( m ) = 5kg
Distance stretched ( Δl ) = 4 * 10^-3 m ( right value )
average bond length ( between atoms ) = 2.3 * 10^-10 m ( right value)
first step : calculate the area
area ( A ) = πr^2 = π * ( 3.5)^2 = 38.48 m^2
γ = MgL / A Δl
= [ (5 * 9.81 * 16 ) / ( 38.48 * (4.3*10^-3) ) ]
= 784.8 / 0.165 = 4756.36 N/m^2
hence : stiffness = γ * bond length
= 4756.36 * 2.3 * 10^-10 = 1.09 * 10^-6 N/m
A person pulls on a 9 kg crate against a 22 Newton frictional force, using a rope attached to the center of the crate. If the The crate began with a speed of 1.5 m/s and speeded up to 2.7 m/s while being pulled a horizontal distance of 2.0 meters. What is the work in J done by the force applied by the rope on the crate
Answer:
uninstell this apps nobody will give u ans its happening to me alsoi was in exam hall i thought this app will give answer but no
An energy efficient light bulb uses 15 W of power for an equivalent light output of a 60 W incandescent light bulb. How much energy is saved each month by using the energy efficient light bulb instead of the incandescent light bulb for 4 hours a day? Assume that there are 30 days in one month
A. 7.2 kW⋅hr
B. 21.6 kW⋅hr
C. 1.8 kW⋅hr
D. 5.4 kW⋅hr
E. 1.35 kW⋅hr
Answer: (d)
Explanation:
Given
15 W is equivalent to 60 W light that is, it save 45 W
So, for 4 hours it is, [tex]4\times 45=180\ W.hr[/tex]
For 30 days, it becomes
[tex]\Rightarrow 180\times 30=5400\ W.hr\\\Rightarrow 5.4\ kWh[/tex]
Thus, [tex]5.4\ kWh[/tex] is saved in 30 days
option (d) is correct.
Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon's tongue accelerates at a remarkable 220 m/s^2 for 20 msms, then travels at constant speed for another 30 ms.
Required:
During this total time of 50 ms, 1/20 of a second, how far does the tongue reach?
Solution :
We know,
Distance,
[tex]$S=ut+\frac{1}{2}at^2$[/tex]
[tex]$S=ut+0.5(a)(t)^2$[/tex]
For the first 20 ms,
[tex]$S=0+0.5(220)(0.020)^2$[/tex]
S = 0.044 m
In the remaining 30 ms, it has constant velocity.
[tex]$v=u+at$[/tex]
[tex]$v=0+(220)(0.020)[/tex]
v = 4.4 m/s
Therefore,
[tex]$S=ut+0.5(a)(t)^2$[/tex]
[tex]$S'=4.4 \times 0.030[/tex]
S' = 0.132 m
So, the required distance is = S + S'
= 0.044 + 0.132
= 0.176 m
Therefore, the tongue can reach = 0.176 m or 17.6 cm
Answer:
The total distance is 0.176 m.
Explanation:
For t = 0 s to t = 20 ms
initial velocity, u = 0
acceleration, a = 220 m/s^2
time, t = 20 ms
Let the final speed is v.
Use first equation of motion
v = u + at
v = 0 + 220 x 0.02 = 4.4 m/s
Let the distance is s.
Use second equation of motion
[tex]s = u t + 0.5 at^2\\\\s = 0 + 0.5 \times 220 \times 0.02\times 0.02\\\\s = 0.044 m[/tex]
Now the distance is
s' = v x t
s' = 4.4 x 0.03 = 0.132 m
The total distance is
S = s + s' = 0.044 + 0.132 = 0.176 m
Two 51 g blocks are held 30 cm above a table. As shown in the figure, one of them is just touching a 30-long spring. The blocks are released at the same time. The block on the left hits the table at exactly the same instant as the block on the right first comes to an instantaneous rest. What is the spring constant?
The concept of this question can be well understood by listing out the parameters given.
The mass of the block = 51 g = 51 × 10⁻³ kgThe distance of the block from the table = 30 cmLength of the spring = 30 cmThe purpose is to determine the spring constant.
Let us assume that the two blocks are Block A and Block B.
At point A on block A, the initial velocity on the block is zero
i.e. u = 0
We want to determine the time it requires for Block A to reach the table. The can be achieved by using the second equation of motion which can be expressed by using the formula.
[tex]\mathsf{S = ut + \dfrac{1}{2}gt^2}[/tex]
From the above formula,
The distance (S) = 30 cm; we need to convert the unit to meter (m).
Since 1 cm = 0.01 mThen, 30cm = 0.3 mThe acceleration (g) due to gravity = 9.8 m/s²
∴
inputting the values into the equation above, we have;
[tex]\mathsf{0.3 = (0)t + \dfrac{1}{2}*(9.80)*(t^2)}[/tex]
[tex]\mathsf{0.3 = \dfrac{1}{2}*(9.80)*(t^2)}[/tex]
[tex]\mathsf{0.3 =4.9*(t^2)}[/tex]
By dividing both sides by 4.9, we have:
[tex]\mathsf{t^2 = \dfrac{0.3}{4.9}}[/tex]
[tex]\mathsf{t^2 = 0.0612}[/tex]
[tex]\mathsf{t = \sqrt{0.0612}}[/tex]
[tex]\mathbf{t =0.247 \ seconds}[/tex]
However, block B comes to an instantaneous rest on point C. This is achieved by the dropping of the block on the spring. During this process, the spring is compressed and it bounces back to oscillate in that manner. The required time needed to get to this point C is half the period, this will eventually lead to the bouncing back of the block with another half of the period, thereby completing a movement of one period.
By applying the equation of the time period of a simple harmonic motion.
[tex]\mathbf{T = 2 \pi \sqrt{\dfrac{m}{k}}}[/tex]
where the relation between time (t) and period (T) is:
[tex]\mathsf{t = \dfrac{T}{2}}[/tex]
T = 2t
T = 2(0.247)
T = 0.494 seconds
[tex]\mathbf{T = 2 \pi \sqrt{\dfrac{m}{k}}}[/tex]
By making the spring constant (k) the subject of the formula:
[tex]\mathbf{\dfrac{T}{2 \pi } = \sqrt{ \dfrac{m}{k}}}[/tex]
[tex]\Big(\dfrac{T}{2 \pi }\Big)^2 = { \dfrac{m}{k}[/tex]
[tex]\dfrac{T^2}{(2 \pi)^2 }= { \dfrac{m}{k}[/tex]
[tex]\mathsf{ T^2 *k = 2 \pi^2*m} \\ \\ \mathsf{ k = \dfrac{2 \pi^2*m}{T^2}}[/tex]
[tex]\mathsf{ k =\Big( \dfrac{(2 \pi)^2*(51 \times 10^{-3})}{(0.494)^2} \Big) N/m}[/tex]
[tex]\mathbf{ k =8.25 \ N/m}[/tex]
Therefore, we can conclude that the spring constant between the two 51 g blocks held at a distance 30 cm above a table as a result of instantaneous rest caused by the compression of the spring is 8.25 N/m.
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Derive the explicit rule for the pattern
3, 0, -3, -6, -9,
A cylindrical swimming pool has a radius 2m and depth 1.3m .it is completely filled with salt water of specific gravity 1.03.The atmospheric preassure is 1.013 x 10^5 Pa.
a.calculate the density of salt water.
Answer:
the density of the salt water is 1030 kg/m³
Explanation:
Given;
radius of the cylindrical pool, r = 2 m
depth of the pool, h = 1.3 m
specific gravity of the salt water, γ = 1.03
The atmospheric pressure, P₀ = 1.013 x 10⁵ Pa
Density of fresh water, [tex]\rho _w[/tex] = 1000 kg/m³
The density of the salt water is calculated as;
[tex]Specific \ gravity \ of \ salt\ water \ (\gamma _s_w) = \frac{density \ of \ salt \ water \ (\rho_{sw})}{density \ of \ fresh \ water \ (\rho_{w})} \\\\1.03 = \frac{\rho_{sw}}{1000 \ kg/m^3}\\\\\rho_{sw} = 1.03 \times 1000 \ kg/m^3\\\\\rho_{sw} = 1030 \ kg/m^3[/tex]
Therefore, the density of the salt water is 1030 kg/m³
A man standing in an elevator holds a spring scale with a load of 5 kg suspended from it. What would be the reading of the scale, if the elevator is accelerating downward with an acceleration 3.8 m/s?.
Answer:
3.1 kg
Explanation:
Applying,
R = m(g-a)..................... Equation 1
Where R = weight of the scale when the elevator is coming down, a = acceleration of the elevator, g = acceleration due to gravith.
From the question,
Given: m = 5 kg, a = 3.8 m/s²
Constant: g = 9.8 m/s²
Substitute these values into equation 1
R = 5(9.8-3.8)
R = 5(6)
R = 30 N
Hence the spring scale is
m' = R/g
m' = 30/9.8
m' = 3.1 kg
A man throw a ball vertically up word with an intial speed 20m/s. What is the maximum height rich by the ball and how long does it take to return to the point it was trow
Answer:
u=20 m/s, T=4s
Explanation:
Given final velocity v= 0 m/s and displacement h= 20 m; acceleration due to gravity = 10 m/ s 2
From equation of motion
v2=u2+2gs−u2=−2(10).20u=20m/s
and time t can be determined by the formula
t=gv−u=−10−20=2s
total time = 2× time of ascend=2×2=4s
it is helpful for you
Liquid plastic is frozen in a physical change that increases its volume. What can be known about the plastic after the change?
(A) Its mass will increase.
(B) Its density will increase.
(C) Its mass will remain the same.
(D) Its density will remain the same.
Answer:
c
Explanation:
Liquid plastic is frozen in a physical change that increases its volume,it can be known about the plastic that Its mass will remain the same, therefore the correct answer is option C.
What is the matter?Anything which has mass and occupies space is known as matter ,mainly there are four states of matter solid liquid gases, and plasma.
These different states of matter have different characteristics according to which they vary their volume and shape.
It is known about plastic that its mass will remain the same when liquid plastic is frozen, by increasing its volume.
Liquid plastic is frozen in a physical change that increases its volume,it can be known about the plastic that Its mass will remain the same, therefore the correct answer is C.
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A parallel plate capacitor creates a uniform electric field of and its plates are separated by . A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed
Complete Question
A parallel plate capacitor creates a uniform electric field of 5 x 10^4 N/C and its plates are separated by 2 x 10^{-3}'m. A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed?
Answer:
[tex]V=1.4*10^5m/s[/tex]
Explanation:
From the question we are told that:
Electric field [tex]B=1.5*10N/C[/tex]
Distance [tex]d=2 x 10^{-3}[/tex]
At negative plate
Generally the equation for Velocity is mathematically given by
[tex]V^2=2as[/tex]
Therefore
[tex]V^2=\frac{2*e_0E*d}{m}[/tex]
[tex]V^2=\frac{2*1.6*10^{-19}(5*10^4)*2 * 10^{-3}}{1.67*10^{-28}}[/tex]
[tex]V=\sqrt{19.2*10^9}[/tex]
[tex]V=1.4*10^5m/s[/tex]
In a mass spectrometer chlorine ions of mass 35u and charge +5e are emitted from a source and accelerated through a potential difference of 250 kV. They then enter a region with a magnetic field that is perpendicular to their original direction of motion. The chlorine ions exit the spectrometer after being bent along a path with radius of curvature 3.5 m. What is the speed of the chlorine ions as they enter the magnetic field region?
(u = 1.66 × 10^(–27) kg, e = 1.6 × 10^(–19) C)
2.6 × 106 m/s
1.2 × 106 m/s
1.5 × 107 m/s
Answer:
v=26.23*105 m/s
or 2.6 × 106 m/s
Explanation:
Force generated by magnetic field will only provide centripetal acceleration thus the entering speed will be same as the exit speed
so,
.5mv2=eV potential differnce*charge= kinetic energy
.5*35*1.66*10-27*v2= 1.6*10-19*5*250000
v2=68.84*1011
v=26.23*105 m/s
or 2.6 × 106 m/s
List what sources of uncertainty go into calculating the wavelength of the laser (no explanation necessary here). (b) Accurately report the uncertainties for these quantities. (c) Explain which of these contributes the most to the final uncertainty on the laser wavelength
Answer:
thanks for da 5points hoi
Explanation: thanks dawg
There can be uncertainty in calculating the wavelength of a laser light due to experimental errors
All measurements have an uncertainty, in the case of direct measurements the uncertainty is equal to the precision of the given instrument.
What are uncertainity in measuring ?Uncertainty means the range of possible values within which the true value of the measurement lies.
What are errors?
The deviation in the value of the measured quantity from the actual quantity or true value is called an error
(a) For the calculation of wavelength of laser light , the sources which can lead to uncertainty are
1. least count of measuring instruments like spectrometer or interferometer
2. Parallax error in the measurement
3. Error in identifying the order of fringes
4.. unable to identify the accurate reading of Vernier or circular scales present in the measuring instruments.
5. Propagating errors
What is least count?
The least count of a measuring instrument is the smallest and accurate value in the measured quantity that can be measured by instrument.
What is propagating error?When you have derived variables, that is, when measurements are made with different instruments, each with a different uncertainty, the way to find the uncertainty or error is that all the errors add up. which increases the uncertainty
b. The uncertainty in measurement due to least count depends on the instrument used for measurement f wavelength. A Michelson's
interferometer has the least count of .0001mm. whereas spectrometer has a least count of 0.5⁰. Hence uncertainty in the measurement by Michelson's interferometer is very less as compared to any other instrument.
C. The maximum uncertainty arises due to the least count , as all other errors can be minimized by taking an average value of many observations but the least count of an instrument do not change so uncertainty within the least count arises.
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Why are hydraulic brakes used?
Answer:
Hydraulic brake systems are used as the main braking system on almost all passenger vehicles and light trucks. Hydraulic brakes use brake fluid to transmit force when the brakes are applied.
Explanation:
Flapping flight is very energy intensive. A wind tunnel test
on an 89 g starling showed that the bird used 12 W of
metabolic power to fly at 11 m/s. What is its metabolic power for starting flight?
Answer:
The metabolic power for starting flight=134.8W/kg
Explanation:
We are given that
Mass of starling, m=89 g=89/1000=0.089 kg
1 kg=1000 g
Power, P=12 W
Speed, v=11 m/s
We have to find the metabolic power for starting flight.
We know that
Metabolic power for starting flight=[tex]\frac{P}{m}[/tex]
Using the formula
Metabolic power for starting flight=[tex]\frac{12}{0.089}[/tex]
Metabolic power for starting flight=134.8W/kg
Hence, the metabolic power for starting flight=134.8W/kg
Se lanza un cohete en un ángulo de 53° sobre la horizontal con una rapidez inicial de 100 m/s. El cohete se mueve por
3.00 s a lo largo de su línea inicial de movimiento con una aceleración de 30.0 m/s2
. En este momento, sus motores fallan,
y el cohete procede a moverse como un proyectil. Determine: (a) la altitud máxima que alcanza el cohete, (b) su tiempo
total de vuelo y (c) su alcance horizontal
Answer:
Explanation:
v = u + at
v₃ = 100 +30.0(3.00) = 190 m/s
s = vt + ½at²
y₃ = (100sin53)(3.00) + ½(30sin53)(3.00²) = 347.4 m
x₃ = (100cos53)(3.00) + ½(30cos53)(3.00²) = 261.8 m
a) v² = u² + uas
s = (v² - u²) / 2a
ymax = 347.4 + (0² - (190sin53)²) / (2(-9.80)) = 1,522 m
b) t₁ = 3.00 s
t₂ = (190sin53) / 9.80 = 15.5 s
t₃ = √(2(1522) / 9.80) = 17.6 s
t = 3.00 + 15.5 + 17.6 = 36.1 s
c) xmax = 261.8 + (190cos53)( 15.5 + 17.6) = 4,047 m
A resistor is submerged in an insulated container of water. A voltage of 12 V is applied to the resistor resulting in a current of 1.2 A. If this voltage and current are maintained for 5 minutes, how much electrical energy is dissipated by the resistor
Explanation:
Given:
[tex]\Delta t = 5\:\text{min} = 300\:\text{s}[/tex]
[tex]V = 12 V[/tex]
[tex]I = 1.2 A[/tex]
Recall that power P is given by
[tex]P = VI[/tex]
so the amount of energy dissipated [tex]\Delta E[/tex] is given by
[tex]\Delta E = VI\Delta t = (12\:\text{V})(1.2\:\text{A})(300\:\text{s})[/tex]
[tex]\:\:\:\:\:\:\:= 4320\:\text{W} = 4.32\:\text{kW}[/tex]
Consider two points in an electric field. The potential at point 1, V1, is 33 V. The potential at point 2, V2, is 175 V. An electron at rest at point 1 is accelerated by the electric field to point 2.
Required:
Write an equation for the change of electric potential energy ΔU of the electron in terms of the symbols given.
Answer:
ΔU = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J
Explanation:
Since the electric potential at point 1 is V₁ = 33 V and the electric potential at point 2 is V₂ = 175 V, when the electron is accelerated from point 1 to point 2, there is a change in electric potential ΔV which is given by ΔV = V₂ - V₁.
Substituting the values of the variables into the equation, we have
ΔV = V₂ - V₁.
ΔV = 175 V - 33 V.
ΔV = 142 V
The change in electric potential energy ΔU = eΔV = e(V₂ - V₁) where e = electron charge = -1.602 × 10⁻¹⁹ C and ΔV = electric potential change from point 1 to point 2 = 142 V.
So, substituting the values of the variables into the equation, we have
ΔU = eΔV
ΔU = eΔV
ΔU = -1.602 × 10⁻¹⁹ C × 142 V
ΔU = -227.484 × 10⁻¹⁹ J
ΔU = -2.27484 × 10⁻²¹ J
ΔU ≅ -2.275 × 10⁻²¹ J
So, the required equation for the electric potential energy change is
ΔU = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J
a stone is thrown vertically upwards with a velocity of 20 m per second what will be its velocity when it reaches a height of 10.2 m
Answer:
Explanation:
Here's the info we have:
initial velocity is 20 m/s;
final velocity is our unknown;
displacement is -10.2 m; and
acceleration due to gravity is -9.8 m/s/s. Using the one-dimensional equation
v² = v₀² + 2aΔx and filling in accordingly to solve for v:
[tex]v=\sqrt{(20)^2+2(-9.8)(-10.2)}[/tex] Rounding to the correct number of sig fig's to simplify:
[tex]v=\sqrt{400+2.0*10^2}[/tex] to get
v = [tex]\sqrt{600}=20\frac{m}{s}[/tex] If you don't round like that, the velocity could be 24, or it could also be 24.5 depending on how your class is paying attention to sig figs or if you are at all.
So either 20 m/s or 24 m/s
If a jet travels 350 m/s, how far will it travel each second?
Answer:
350
Explanation:
Since it travels 350 meters per second, the jet will travel 350 meters in one second.
An object whose weight is 100 lbf experiences a decrease in kinetic energy of 500 ft lbf and an increase in potential energy of 1500 ft lbf. The initial velocity and elevation of the object, each relative to the surface of the earth, are 40 ft/s and 30 ft, respectively. If g 5 32.2 ft/s2 , determine:
(a) the final velocity, in ft/s.
(b) the final elevation, in ft.
Answer:
a) [tex]v_2=35.60ft/sec[/tex]
b) [tex]h_2=45ft[/tex]
Explanation:
From the question we are told that:
Weight [tex]W=100lbf[/tex]
Decrease in kinetic energy [tex]dK.E= 500 ft lbf[/tex]
Increase in potential energy [tex]dP.E =1500 ft lbf.[/tex]
Velocity [tex]V_1=40[/tex]
Elevation [tex]h=30ft[/tex]
[tex]g=32.2 ft/s2[/tex]
a)
Generally the equation for Change in Kinetic Energy is mathematically given by
[tex]dK.E=\frac{1}{2}m(v_1^2-v_2^2)[/tex]
[tex]500=\frac{1}{2}*\frac{100}{32.2}(v_1^2-v_2^2)[/tex]
[tex]v_2=35.60ft/sec[/tex]
b)
Generally the equation for Change in Potential Energy is mathematically given by
[tex]dP.E=mg(h_2-h_1)[/tex]
[tex]1500=mg(h_2-h_1)[/tex]
[tex]h_2=45ft[/tex]
The 52-g arrow is launched so that it hits and embeds in a 1.50 kg block. The block hangs from strings. After the arrow joins the block, they swing up so that they are 0.47 m higher than the block's starting point. How fast was the arrow moving before it joined the block? What mechanical work must you do to lift a uniform log that is 3.1 m long and has a mass of 100 kg from the horizontal to a vertical position?
Answer:
[tex]v_1=87.40m/s[/tex]
Explanation:
From the question we are told that:
Mass of arrow [tex]m=52g[/tex]
Mass of rock [tex]m_r=1.50kg[/tex]
Height [tex]h=0.47m[/tex]
Generally the equation for Velocity is mathematically given by
[tex]v = \sqrt{(2gh)}[/tex]
[tex]v=\sqrt{(2 * 9.8m/s² * 0.47m) }[/tex]
[tex]v= 3.035m/s[/tex]
Generally the equation for conservation of momentum is mathematically given by
[tex]m_1v_1=m_2v_2[/tex]
[tex]0.052kg * v = 1.5 * 3.03m/s[/tex]
[tex]v_1=87.40m/s[/tex]
You drive 7.5 km in a straight line in a direction east of north.
a. Find the distances you would have to drive straight east and then straight north to arrive at the same point.
b. Show that you still arrive at the same point if the east and north legs are reversed in order.
Answer:
a) a = 5.3 km, b) sum fulfills the commutative property
Explanation:
This is a vector exercise, If you drive east from north, we can find the vector using the Pythagorean theorem
R² = a² + b²
where R is the resultant vector R = 7.5 km and the others are the legs
If we assume that the two legs are equal to = be
R² = 2 a²
r = √2 a
a = r /√2
we calculate
a = 7.5 /√2
a = 5.3 km
therefore, you must drive 5.3 km east and then 5.3 km north and you will reach the same point
b) As the sum fulfills the commutative property, the order of the elements does not alter the result
a + b = b + a
therefore, it does not matter in what order the path is carried out, it always reaches the same end point
A block of mass 0.260 kg is placed on top of a light, vertical spring of force constant 5 200 N/m and pushed downward so that the spring is compressed by 0.090 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise
After being released, the restoring force exerted by the spring performs
1/2 (5200 N/m) (0.090 m)² = 12.06 J
of work on the block. At the same time, the block's weight performs
- (0.260 kg) g (0.090 m) ≈ -0.229 J
of work. Then the total work done on the block is about
W ≈ 11.83 J
The block accelerates to a speed v such that, by the work-energy theorem,
W = ∆K ==> 11.83 J = 1/2 (0.260 kg) v ² ==> v ≈ 9.54 m/s
Past the equilibrium point, the spring no longer exerts a force on the block, and the only force acting on it is due to its weight, hence it has a downward acceleration of magnitude g. At its highest point, the block has zero velocity, so that
0² - v ² = -2gy
where y is the maximum height. Solving for y gives
y = v ²/(2g) ≈ 4.64 m
