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ELEN [110]
3 years ago
7

A particular type of fundamental particle decays by transforming into an electron e- and a positron e (same amount of charge as

e- but with positive sign). Suppose the decaying particle is at rest in a uniform magnetic field B of magnitude 3.53 mT and the e- and e move away from the decay point in paths lying in a plane perpendicular to B. How long after the decay do the e- and e collide
Physics
1 answer:
Solnce55 [7]3 years ago
7 0

Answer:

Time after which it collide again is

T = 1.01 \times 10^{-8} s

Explanation:

As we know that the neutral particle decays into two particles i.e. electron and positron

As we know that both particles are of same mass but opposite charge

So both particles will revolve in the circle in opposite sense

So both particles will collide again after one complete revolution

So the time period of revolution is given as

T = \frac{2\pi m}{qB}

so we whave

T = \frac{2\pi(9.1 \times 10^{-31})}{(1.6 \times 10^{-19})(3.53 \times 10^{-3})}

T = 1.01 \times 10^{-8} s

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Work package. Hope this helps!
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3 years ago
A pulley lifts a 72-N load with a force of 24-N. The input distance is 2m and the output distance is 0.5m. What is the efficienc
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Answer:

Explanation:

Work done on the lever ( input energy ) = force applied x input distance

= 24 N x 2m = 48 J

Work done by the lever ( output energy ) = load x output distance

= 72 N x 0.5m = 36 J

efficiency = output energy / input energy

= 36 J  / 48 J

= 3 / 4 = .75

In percentage terms efficiency = 75 % .

5 0
3 years ago
Three observers watch a train pull away from a station toward the right of the platform. Observer A is in one of the train’s car
juin [17]

Observer A is moving inside the train

so here observer A will not be able to see the change in position of train as he is standing in the same reference frame

So here as per observer A the train will remain at rest and its not moving at all

Observer B is standing on the platform so here it is a stationary reference frame which is outside the moving body

So here observer B will see the actual motion of train which is moving in forward direction away from the platform

Observer C is inside other train which is moving in opposite direction on parallel track. So as per observer C the train is coming nearer to him at faster speed then the actual speed because they are moving in opposite direction

So the distance between them will decrease at faster rate

Now as per Newton's II law

F = ma

Now if train apply the brakes the net force on it will be opposite to its motion

So we can say

- F = ma

a = \frac{-F}{m}

so here acceleration negative will show that train will get slower and its distance with respect to us is now increasing with less rate

It is not affected by the gravity  because the gravity will cause the weight of train and this weight is always counterbalanced by normal force on the train

So there is no effect on train motion



5 0
3 years ago
g A bowling ball with a mass of 3.86 kg and a radius of 0.161 m starts from rest at a height of 2.5 m and rolls down a 48.4 o sl
Fynjy0 [20]

Answer:

v=1.5m/s

Explanation:

The gravitational potential energy gets transformed into translational and rotational kinetic energy, so we can write mgh=\frac{mv^2}{2}+\frac{I\omega^2}{2}. Since v=r\omega (the ball rolls without slipping) and for a solid sphere I=\frac{2mr^2}{5}, we have:

mgh=\frac{mv^2}{2}+\frac{2mr^2\omega^2}{2*5}=\frac{mv^2}{2}+\frac{mv^2}{5}=\frac{7mv^2}{10}

So our translational speed will be:

v=\sqrt{\frac{10gh}{7}}=\sqrt{\frac{10(9.8m/s^2)(0.161m)}{7}}=1.5m/s

6 0
2 years ago
When we come out in the sun, we feel warm. How does the heat from the sun reach us?
dsp73

Answer:

<h2>Radiation</h2>

Explanation:

<h3>Greetings !</h3>

The Sun reaches us by propagating through the vacuum of space. Sunlight reaches the Earth in about 8 minutes and 20 seconds. When this energy reaches the Earth's atmosphere, both conduction and convection play key roles to scatter it throughout the planet.

3 0
1 year ago
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