Ask question

Ask question

All categories

3 years ago

15

A charge partides round a 1 m radius circular particle accelerator at nearly the speed of light. Find : (a) The period (b) The c

entripetal acceleration of the charged particles

1 answer:

Scrat [10]3 years ago

8 0

Explanation:

It is given that,

Radius of circular particle accelerator, r = 1 m

The distance covered by the particle is equal to the circumference of the circular path, d = 2πr

d = 2π × 1 m

(a) The speed of satellite is given by total distance divided by total time taken as :

$speed=\dfrac{distance}{time}$

Let t is the period of the particle.

$t=\dfrac{d}{s}$

d = distance covered

s = speed of particle

It is given that the charged particle is moving nearly with the speed of light

$t=\dfrac{d}{c}$

$t=\dfrac{2\pi\times 1\ m}{3\times 10^8\ m/s}$

$t=2.09\times 10^{-8}\ s$

(b) On the circular path, the centripetal acceleration is given by :

$a=\dfrac{c^2}{r}$

$a=\dfrac{(3\times 10^8\ m/s)^2}{1\ m}$

$a=9\times 10^{16}\ m/s^2$

Hence, this is the required solution.

You might be interested in

The deviation of a value (x) from the mean of the sample divided by the standard deviation (s)of the sample is known as the "z"

Ugo [173]

Answer:

true.

Explanation:

z-score gives an idea of how far the value of data is from the mean point. Basically is the measure of a standard deviation below or above the mean.

More the deviation of the data from the mean more is the standard deviation.

Z-score or value is calculated by the expression given below:

$z = \dfrac{x-\mu}{\sigma}$

x is the value

μ is the mean

σ is the standard deviation

Hence, the statement given is true.

6 0

3 years ago

An object is dropped from a bridge. A second object is thrown downwards 1.0 s later. They both reach the water 20 m below at the

olga2289 [7]

Answer:

vi = 14.610

Explanation:

initial velocity of the first object (vi) = 0 m/s, because it was dropped

distance (y) = -20 m

initial position (y0) = 0 m

acceleration due to gravity (g) = -9.8 m/s^2

y = 1/2 gt^2 + vi*t + y0

-20 = 1/2(-9.8)t^2 + 0 + 0

-20 = -4.9t^2

4.081 = t^2

+√4.081 = t

t = 2.020

time of second object = 2.020 - 1 = 1.020

Now we can plug in the new time to solve for vi of the second object.

y = 1/2 gt^2 + vi*t + y0

-20 = 1/2(-9.8)1.020^2 + vi*1.020 + 0

-20 = -5.098 + 1.020vi

14.902 = 1.020vi

vi = 14.610 m/s

6 0

3 years ago

An object of mass 2 kg travels through outer space in a straight line at a constant

slava [35]

Answer:

b. It is subject to a net force of zero

Explanation:

F = m*a

F = force

m = mass

a = acceleration

Since the object is moving at a constant velocity not speeding up or slowing down and moving in a sight line. All the forces that are acting on the object are canceling each other out so the answer is:

b. It is subject to a net force of zero

7 0

3 years ago

Consider a ship of mass 2 x 10^6 kg with propeller blades of radius 7 m and it can push water backwards to a speed of 2 m/s. Giv

PolarNik [594]

Answer:

$a = 0.308 m/s^2$

Explanation:

As we know that the force due to propeller when it will push the water backwards is given as

$F = \frac{dP}{dt}$

now we know that

$P = mv$

so we have

$F = v\frac{dm}{dt}$

$F = v \rhoA\frac{dx}{dt}$

$F = \rho A v^2$

here we know that

$A = \pi r^2$

$A = \pi(7^2)$

$A = 154 m^2$

now the force is given as

$F = (1000)(154)(2^2)$

$F = 616000 N$

now the acceleration of the ship is given by Newton's II law

$a = \frac{F}{m}$

$a = \frac{616000}{2 \times 10^6}$

$a = 0.308 m/s^2$

3 0

3 years ago

Are gases more or less dense than liquids or solids at room temperature?

tigry1 [53]

<span>Solids are denser than liquids, and liquids are denser than gases at room temperature. </span>

8 0

4 years ago