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inn [45]
2 years ago
14

Imagine that you are sitting in a closed room (no windows, no doors) when, magically, it is lifted from Earth and sent accelerat

ing through space with an acceleration of 1g (9.8 m/s2). According to Einstein's equivalence principle, which of the following is true?
You won't feel any change and will have no way to know that you've left the earth.
You'll feel a force that will cause your head to repeatedly bang into the ceiling.
You'll know that you left the Earth because you'll be floating weightlessly in your room.
You'll know that you left the Earth because when you drop a ball it will fall sideways.
Physics
1 answer:
Alik [6]2 years ago
7 0

Answer:

You won't feel any change and will have no way to know that you've left the earth.

Explanation:

One of the formulations of the principle says that the properties of an inertial system under the effect of a gravitational field are the same as a non-inertial(accelerated) system. The inertial system under the effect of a gravitational field occurs when you are in the room on earth, and when the room is sent accelerating through space you have the non-inertial system. Both have the same properties, so it's impossible to tell the difference if you can't look outside.

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2 years ago
If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g
yulyashka [42]

Answer:

178200 g mile pounds

Explanation:

Work= Force * Distance= Fh

F=ma=mg where m is mass and g is acceleration due to gravity

Work= 165 pounds *g* 1080 m=  178200 g mile pounds

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2 years ago
A seasoned mini golfer is trying to make par on a tricky number five hole. The golfer can complete the hole by hitting the ball
liberstina [14]

Answer:5.17 m/s

Explanation:

Given

let u be the speed at cliff initial point

range over cliff is 1.45 m

and range of projectile is given by

R=\frac{u^2\sin 2\theta }{g}

1.45=\frac{u^2\sin 90}{9.8}

u=3.77 m/s

Conserving Energy

E_{bottom}=E_{initial\ point\ at\ cliff}

Kinetic energy=Kinetic energy +Potential energy gained

Let v be the initial velocity

\frac{mv^2}{2}=mgh+\frac{mu^2}{2}

v^2=u^2+2gh

v=\sqrt{u^2+2gh}

v=\sqrt{3.77^2+2\time 9.8\times 0.64}

v=\sqrt{26.75}=5.17 m/s

5 0
1 year ago
The mass per unit length of a 14-gauge copper wire is 18.5 g/m. If the wire is placed running along the horizontal x-axis (east-
zmey [24]

Answer:

0.6295 A

Explanation:

I=mg/BL put values in this formula.  

7 0
1 year ago
A turntable rotates counterclockwise at 76 rpm . A speck of dust on the turntable is at 0.47 rad at t=0. What is the angle of th
icang [17]

To solve this exercise it is necessary to apply the kinematic equations of angular motion.

By definition we know that the displacement when there is constant angular velocity is

\theta= \theta_0 +\omega t

From our given data we know that,

\omega = 76\frac{rev}{min}

\omega = 76\frac{rev}{min}(\frac{2\pi rad}{1rev})(\frac{1 min}{60s})

\omega = 7.958rad/s

Moreover we know that

\theta_0 = 0.47 rad

Therefore for time t=8.1s we have,

\theta= \theta_0+ \omega t

\theta= 0.47+(7.958)(8.1)

\theta = 64.9298rad

That number in revolution is:

\theta = 64.9298rad(\frac{1rev}{2\pi})

\theta = 15.108 Revolutions

Here, we see that there are 15 complete revolutions

And 0.108 revolutions i not complete, so the tunable rotation is

\theta_{net} = 0.108*2\pi=0.216\pi

Therefore the angle of the speck at a time 8.1s is 0.216\pi

4 0
2 years ago
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