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3 years ago

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Consider an electron moving with relativistic speed v and momentum of magnitude p. If the speed of the electron is doubled, the

resulting relativistic momentum of the electron will be:a) equal to 2pb) less than 2pc) greater than 2pPlease explain why.

1 answer:

astraxan [27]3 years ago

8 0

Answer:

option (b)

Explanation:

initial speed = v

initial momentum = p

final speed = 2v

final momentum = p'

The formula for relativistic momentum is given by

$p = \frac{mv}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$ .... (1)

The new momentum is given by

$p' = \frac{m\times 2v}{\sqrt{1-\frac{4v^{2}}{c^{2}}}}$ .... (2)

By comparison, p' < 2p.

Thus, the option is (b) .

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A 1.3-kg model airplane flies in a circular path on the end of a 23-m line. The plane makes

storchak [24]

(a) The plane makes 4.3 revolutions per minute, so it makes a single revolution in

(1 min) / (4.3 rev) ≈ 0.2326 min ≈ 13.95 s ≈ 14 s

(b) The plane completes 1 revolution in about 14 s, so that in this time it travels a distance equal to the circumference of the path:

(2<em>π</em> (23 m)) / (14 s) ≈ 10.3568 m/s ≈ 10 m/s

(c) The plane accelerates toward the center of the path with magnitude

<em>a</em> = (10 m/s)² / (23 m) ≈ 4.6636 m/s² ≈ 4.7 m/s²

(d) By Newton's second law, the tension in the line is

<em>F</em> = (1.3 kg) (4.7 m/s²) ≈ 6.0627 N ≈ 6.1 N

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3 years ago

An electron is released from rest in a weak electric field given by E =-2.30 x 10-10 N/Cj. After the electron has traveled a ver

Goshia [24]

Explanation:

It is given that,

An electron is released from rest in a weak electric field of, $E=2.3\times 10^{-10}\ N/C$

Vertical distance covered, $s=1\ \mu m=10^{-6}\ m$

We need to find the speed of the electron. Let its speed is v. Using third equation of motion as :

$v^2-u^2=2as$

$v^2=2as$ .............(1)

Electric force is $F_e$ and force of gravity is $F_g$ . As both forces are acting in downward direction. So, total force is:

$F=mg+qE$

$F=9.1\times 10^{-31}\times 9.8+1.6\times 10^{-19}\times 2.3\times 10^{-10}$

$F=4.57\times 10^{-29}\ N$

Acceleration of the electron, $a=\dfrac{F}{m}$

$a=\dfrac{4.57\times 10^{-29}\ N}{9.1\times 10^{-31}\ kg}$

$a=50.21\ m/s^2$

Put the value of a in equation (1) as :

$v=\sqrt{2as}$

$v=\sqrt{2\times 50.21\times 10^{-6}}$

v = 0.010 m/s

So, the speed of the electron is 0.010 m/s. Hence, this is the required solution.

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3 years ago

A bird is traveling west with a velocity of 8 m/s and momentum of 40 mg*m/s what's the birds mass?

VikaD [51]

Answer:

m = 5 [mg]

Explanation:

We must remember that the definition of linear momemtum is defined as the product of mass by distance.

P = m*v

P = momentum = 40 [mg*m/s]

m = mass [mg]

v = velocity = 8 [m/s]

Now clearing m:

m = P/v

m = 40/8

m = 5 [mg]

3 0

3 years ago

Your 64-cm-diameter car tire is rotating at 3.3 rev/swhen suddenly you press down hard on the accelerator. After traveling 250 m

umka21 [38]

Answer:

0.76 rad/s^2

Explanation:

First, we convert the original and final velocity from rev/s to rad/s:

$v_o = 3.3\frac{rev}{s} * \frac{2\pi rad}{1rev} =20.73 rad/s$

$v_f = 6.4\frac{rev}{s} * \frac{2\pi rad}{1rev}=40.21 rad/s$

Now, we need to find the number of rads that the tire rotates in the 250m path. We use the arc length formula:

$D = x*r \\x = \frac{D}{r} = \frac{250m}{0.64m/2} = 781.25 rads$

Now, we just use the formula:

$w_f^2-w_o^2=2\alpha*x$

$\alpha =\frac{w_f^2-w_o^2}{2x} = \frac{(40.21rad/s)^2-(20.73rad/s)^2}{2*781.25rad} = 0.76 rad/s^2$

6 0

2 years ago

Read 2 more answers

you stop the stopwatch at 4.0 s, but you notice a short time later that the same ant is at 0.81 m on the meter stick. Assuming t

telo118 [61]

The time elapsed since you stopped the stopwatch is 0.41 s.

<em>Your question is not complete, it seems to be missing the following information;</em>

"The velocity of the ant is 2 m/s"

The given parameters;

velocity of the ant, v = 2 m/s

change in position of the ant, Δx = 0.81 m

initial time, t₁ = 4 s

time when the ant was noticed, = t₂

Velocity is defined as the change in displacement per change in time of motion of an object.

$v = \frac{\Delta x}{\Delta t} = \frac{\Delta x}{t_2 - t_1} \\\\t_2 -t_1 = \frac{\Delta x}{v} \\\\t_2 - 4 = \frac{0.81}{2} \\\\t_2 - 4 = 0.405\\\\t_2 = 0.405 + 4\\\\t_2 = 4.405 \approx 4.41 \ s$

The time elapsed since you stopped the stopwatch is calculated as;

$t_{elapsed} = 4.41 \ s - 4\ s = 0.41 \ s$

Thus, the time elapsed since you stopped the stopwatch is 0.41 s.

Learn more here: brainly.com/question/18153640

5 0

3 years ago