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3 years ago

When released , what is the kinetic energy of the 1c charge of the preceding problem if it flies past its starting position?

1 answer:

5 0

A) When a charge is moved in an electric field the work done (W) is calculated as charge*(change in potential). We can write W = q*V or V = W/q = 10/1 = 10V . This voltage is a difference in electric potential between 2 points within the field. If the charge is positive, and positive work is done upon it, then the final position is more positive than the original one.

<span>b) If a charge (Q) is released from rest and falls through a potential difference V, then its gain in energy (KE if no other force acts on the charged body) is q*V = 10J. This is the same as the work done in moving the charge to its new position in part (a), and is an example of the conservation of energy.</span>

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A train travels due south at 25 m/s (relative to the ground) in a rain that is blown toward the south by the wind. The path of e

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Answer:

Explanation:

Given

Train travels towards south with a velocity if $v_t=25\ m/s$

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suppose $v_r$ is the velocity of rain with respect to ground then

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$v_r\times \sin (66)=25$

$v_r=27.36\ m/s$

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A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf

stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

$F - mg sin \theta = 0$

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=30.0^{\circ}$ is the angle of the incline

Solving for F,

$F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N$

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

$P=Fv = (4655)(2.20)=10241 W$

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

$v=u+at$

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

$a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2$

So now the equation of the forces along the direction parallel to the incline is

$F - mg sin \theta = ma$

And solving for F, we find the maximum force applied by the motor:

$F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N$

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

$P=Fv=(4829)(2.20)=10624 W$

(c) $5.82\cdot 10^6 J$

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

$\Delta U = mg \Delta h$