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3 years ago

8

If the strength of the magnetic field at A is 48 units and the strength of the magnetic field at B is 3 units, what is the dista

nce to B? (Drawing is not to scale!)

1 answer:

lilavasa [31]3 years ago

8 0

This is the formula to calculate the distance:
Force (F)= charge (q) * velocity (v) * field strength (B)

In the end, you get 20 units, which should be the correct answer.

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A 0.300kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.800m/s when it makes a head-o

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Answer:

The final velocity of the first glider is 0.27 m/s in the same direction as the first glider

The final velocity of the second glider is 1.07 m/s in the same direction as the first glider.

0.010935 J

0.0858675 J

Explanation:

$m_1$ = Mass of first glider = 0.3 kg

$m_2$ = Mass of second glider = 0.15 kg

$u_1$ = Initial Velocity of first glider = 0.8 m/s

$u_2$ = Initial Velocity of second glider = 0 m/s

$v_1$ = Final Velocity of first glider

$v_2$ = Final Velocity of second glider

As momentum and Energy is conserved

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

${\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}={\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}$

From the two equations we get

$v_{1}=\frac{m_1-m_2}{m_1+m_2}u_{1}+\frac{2m_2}{m_1+m_2}u_2\\\Rightarrow v_1=\frac{0.3-0.15}{0.3+0.15}\times 0.8+\frac{2\times 0.15}{0.3+0.15}\times 0\\\Rightarrow v_1=0.27\ m/s$

The final velocity of the first glider is 0.27 m/s in the same direction as the first glider

$v_{2}=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 0.3}{0.3+0.15}\times 0.8+\frac{0.3-0.15}{0.3+0.15}\times 0\\\Rightarrow v_2=1.067\ m/s$

The final velocity of the second glider is 1.07 m/s in the same direction as the first glider.

Kinetic energy is given by

$K=\frac{1}{2}m_1v_1^2\\\Rightarrow K=\frac{1}{2}0.3\times 0.27^2\\\Rightarrow K=0.010935\ J$

Final kinetic energy of first glider is 0.010935 J

$K=\frac{1}{2}m_2v_2^2\\\Rightarrow K=\frac{1}{2}0.15\times 1.07^2\\\Rightarrow K=0.0858675\ J$

Final kinetic energy of second glider is 0.0858675 J

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A cyclotron (Fig. 29.16) designed to accelerate protons has an outer radius of 0.350 m . The protons are emitted nearly at rest

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The time interval for which the proton accelerated can be calculated using -

$$t=\frac{N}{f}=$ $$\frac{3.3\times 10^{25} }{f}$

Using the above formula you can find the time interval for any frequency.

We have a Cyclotron designed to accelerate protons.

We have to determine for what time interval does the proton accelerate.

<h3>What is a Cyclotron? On what principle it works?</h3>

Cyclotron is a device used to accelerate charged particles to high energies. It works on the principle that a charged particle moving normal to a magnetic field experiences magnetic Lorentz force due to which the particle moves in a circular path. The magnetic Lorentz force is given by-

F = qvB sinθ

According to the question -

In a cyclotron the force by a magnetic field is equal to the centrifugal force.

Now, for r(max) - ( Since v = rω)

q v(max) B sin (90) = $$\frac{m\times v(max)^{2} }{r(max)}$

$$v(max)=\frac{qBr(max)}{m}$

Now -

r(max) = 0.35 m

B = 0.8 Tesla

Therefore -

v(max) = $$\frac{1.6\times 10^{-19} \times 0.8 \times 0.35}{1.6\times 10^{-27} }$

v(max) = 0.28 x $10^{8}$ m/sec

Therefore, the Maximum kinetic energy = Kinetic energy at which they leave the cyclotron-

E(max) = $\frac{1}{2} m\times v(max)^{2}$

E(max) = $\frac{1}{2} \times 1.6\times 10^{-27} \times 0.28\times 10^{8} \times 0.28\times 10^{8}$

E(max) = 0.063 x $10^{11}$ joules.

Now -

The kinetic energy K is built up from 2N passes through Voltage V = 600 volts. Therefore -

$$N = \frac{E(max)}{2qV}$ = $$\frac{0.063\times 10^{11} }{2\times 1.6\times 10^{-19} \times 600}$ = 3.3 x $10^{25}$ revolutions.

The total number of revolutions are 3.3 x $10^{25}$ revolutions.

Now - The time interval for which the proton accelerated can be calculated using -

$$t=\frac{N}{f}=$ $$\frac{3.3\times 10^{25} }{f}$

Using the above formula you can find the time interval for any frequency.

To solve more questions on Cyclotron, visit the link below-

brainly.com/question/14555284

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