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3 years ago

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A charge is accelerated from rest through a potential difference V and then enters a uniform magnetic field oriented perpendicul

ar to its path. The field deflects the particle into a circular arc of radius R. If the accelerating potential is tripled to 3V, what will be the radius of the circular

1 answer:

lys-0071 [83]3 years ago

4 0

Answer:

$\sqrt{3}R$

Explanation:

Case 1:

$\Delta V$ = potential difference through which the charge is accelerated = V

$m$ = mass of the charge

$q$ = magnitude of charge

$v$ = speed gained by the charge due to potential difference

Using conservation of energy

Kinetic energy gained = Electric potential energy lost

$(0.5)mv^{2}=q\Delta V$

$v = \sqrt{\frac{2q\Delta V}{m}}$

$B$ = magnitude of magnetic field

$R$ = Radius of arc

Radius of arc is given as

$R = \frac{mv}{qB}$

$R = \frac{m}{qB}\sqrt{\frac{2q\Delta V}{m}}$

$R = \frac{1}{B}\sqrt{\frac{2m V}{q}}$ eq-1

Case 2:

$\Delta V'$ = potential difference through which the charge is accelerated = 3V

$m$ = mass of the charge

$q$ = magnitude of charge

$v'$ = speed gained by the charge due to potential difference

Using conservation of energy

Kinetic energy gained = Electric potential energy lost

$(0.5)mv'^{2}=q\Delta V'$

$v' = \sqrt{\frac{2q\Delta V'}{m}}$

$B$ = magnitude of magnetic field

$R'$ = Radius of arc

Radius of arc is given as

$R' = \frac{mv'}{qB}$

$R' = \frac{m}{qB}\sqrt{\frac{2q\Delta V'}{m}}$

$R' = \frac{1}{B}\sqrt{\frac{2m\Delta V'}{q}}$

$R' = \frac{1}{B}\sqrt{\frac{2m (3V)}{q}}$

$R' = \sqrt{3}\frac{1}{B}\sqrt{\frac{2m V}{q}}$

Using eq-1

$R' = \sqrt{3}R$

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A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r

atroni [7]

(a) $2.79 rev/s^2$

The angular acceleration can be calculated by using the following equation:

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where:

$\omega_f = 20.0 rev/s$ is the final angular speed

$\omega_i = 11.0 rev/s$ is the initial angular speed

$\alpha$ is the angular acceleration

$\theta=50.0 rev$ is the number of revolutions made by the disk while accelerating

Solving the equation for $\alpha$ , we find

$\alpha=\frac{\omega_f^2-\omega_i^2}{2d}=\frac{(20.0 rev/s)^2-(11.0 rev/s)^2}{2(50.0 rev)}=2.79 rev/s^2$

(b) 3.23 s

The time needed to complete the 50.0 revolutions can be found by using the equation:

$\alpha = \frac{\omega_f-\omega_i}{t}$

where

$\omega_f = 20.0 rev/s$ is the final angular speed

$\omega_i = 11.0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

t is the time

Solving for t, we find

$t=\frac{\omega_f-\omega_i}{\alpha}=\frac{20.0 rev/s-11.0 rev/s}{2.79 rev/s^2}=3.23 s$

(c) 3.94 s

Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

$\alpha = \frac{\omega_f-\omega_i}{t}$

where

$\omega_f = 11.0 rev/s$ is the final angular speed

$\omega_i = 0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

t is the time

Solving for t, we find

$t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s$

(d) 21.7 revolutions

The number of revolutions made by the disk to reach the 11.0 rev/s angular speed can be found by using

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where:

$\omega_f = 11.0 rev/s$ is the final angular speed

$\omega_i = 0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

$\theta=?$ is the number of revolutions made by the disk while accelerating

Solving the equation for $\theta$ , we find

$\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev$

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3 years ago

2. A 2000 kg car with speed 12.0 m/s hits a tree. The tree does not move or

krek1111 [17]

a) The work done by the tree is $-1.44\cdot 10^5 J$

b) The amount of force applied is 2880 N

Explanation:

a)

According to the work-energy theorem, the work done on the car is equal to the change in kinetic energy of the car. Therefore, we can write:

$W=K_f - K_i = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

where

W is the work done on the car

m is the mass of the car

u is its initial speed

v is its final speed

For the car in this problem, we have:

m = 2000 kg

u = 12.0 m/s

v = 0 (since the car comes to a stop, after the crash)

Therefore, the work done by the tree on the car is:

$W=0-\frac{1}{2}(2000)(12.0)^2=-1.44\cdot 10^5 J$

The work is negative because it is done in the direction opposite to the direction of motion of the car.

b)

The work done by the tree on the car can also be rewritten as

$W=Fd$

where

F is the force applied on the car

d is the displacement of the car during the collision

In this situation, we have:

$W=-1.44\cdot 10^5 J$ is the work done

$d=50.0 cm = 0.50 m$ is the displacement of the car during the collision

Solving the equation for F, we find the force exerted by the tree on the car:

$F=\frac{W}{d}=\frac{-1.44\cdot 10^5 J}{0.50}=-2880 N$

Where the negative sign means the force is applied opposite to the direction of motion of the car. Therefore, the magnitude of the force applied is 2880 N.

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A billiard ball (ball #1) moving at 5.00 m/s strikes a stationary ball (ball #2) of the same mass. after the collision, ball #1

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If it is completely elastic, you can calculate the velocity of the second ball from the kinetic energy
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<span>kinetic energy: v1^2 = v1' ^2 + v2' ^2 (1/2 and m cancel out) </span>
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