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Bumek [7]
3 years ago
13

A crew on a spacecraft watches a movie that is two hours long. The spacecraft is moving at high speed through space. An Earth-ba

sed observer watching the movie screen on the spacecraft through a powerful telescope measures the duration of the movie to be _______ two hours.
Physics
1 answer:
laiz [17]3 years ago
4 0

Answer:

The duration of the movie is longer than 2 hrs.

Explanation:

Given:

The duration of the movie observed by the crew on the spacecraft is 2 hrs.

According to time-dilation formula:

t = \dfrac{t_{0}}{\sqrt{1 - \dfrac{v^{2}}{c^{2}}}}

Here, t is the required time, t_{0} is the original time, v is the velocity of the spacecraft and c is the velocity of light.

Since v, so t_{0} > t.

So the time required will be large.

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The speed of an object changes only when it is acted on by an unbalanced force.
Nikitich [7]

Answer:

If an object has a net force acting on it, it will accelerate. The object will speed up, slow down or change direction. An unbalanced force (net force) acting on an object changes its speed and/or direction of motion. An unbalanced force is an unopposed force that causes a change in motion.

Explanation:

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6 0
3 years ago
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Answer:

Explanation:

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3 years ago
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A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5×10−3m2. The solenoid is 0.50 m long and has 500
defon

To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.

This definition is described in the following equation as,

M = \frac{\mu_0 N_1 N_2A_1}{l_1}

Where,

\mu =permeability of free space

N_1 = Number of turns in solenoid 1

N_2 = Number of turns in solenoid 2

A_1= Cross sectional area of solenoid

l = Length of the solenoid

Part A )

Our values are given as,

\mu_0 = 4\pi *10^{-7}H/m

N_1 = 500

N_2 = 40

A = 7.5*10^{-4}m^2

l = 0.5m

Substituting,

M = \frac{\mu_0 N_1 N_2A_2}{l_1}

M = \frac{(4\pi *10^{-7})(500)(40)(7.5*10^{-4})}{0.5}

M = 3.77*10^{-4}H

PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.

8 0
3 years ago
Many industries are powered via distant power stations. Calculate the current flowing through a 7,300m long 10. copper power lin
Oliga [24]

Answer:

Current, I = 1000 A

Explanation:

It is given that,

Length of the copper wire, l = 7300 m

Resistance of copper line, R = 10 ohms

Magnetic field, B = 0.1 T

\mu_o=4\pi \times 10^{-7}\ T-m/A

Resistivity, \rho=1.72\times 10^{-8}\ \Omega-m

We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :

R=\rho \dfrac{l}{A}

R=\rho \dfrac{l}{\pi r^2}

r=\sqrt{\dfrac{\rho l}{R\pi}}

r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}

r = 0.00199 m

or

r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m

The magnetic field on a current carrying wire is given by :

B=\dfrac{\mu_o I}{2\pi r}

I=\dfrac{2\pi rB}{\mu_o}

I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

4 0
3 years ago
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