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3 years ago

Why to astronauts appear weightless while they are filmed performing activities inside the orbiting space shuttle?

Physics

2 answers:

vagabundo [1.1K]3 years ago

8 0

<h2>Answer: The astronauts are falling at the same rate as the space shuttle as it orbits around earth</h2>

The astronauts seem to float because they are in free fall just like the spacecraft.

However, although they are constantly falling on the Earth, they do not fall because the ship orbits at a sufficient speed (in the same direction of rotation of the Earth) so that the centrifugal force is balanced with the Earth's gravitational pull.

In other words:

The spaccraft and the astronauts are in free fall but the Earth's surface will never be reached as long as they does not decrease the speed.

Then, as they accelerate toward Earth (regardless of their mass), it curves beneath them and never comes close.

That's why astronauts, having the same acceleration as the spacecraft, feel weightless and see themselves floating.

Alexeev081 [22]3 years ago

5 0

Answer:

an astronaut is under the gravitational force of any object nearby that has mass.

Explanation:

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A tennis racket hits a tennis ball with a force of F=at−bt2, where a = 1290 N/ms , b = 330 N/ms2 , and t is the time (in millise

denpristay [2]

Answer:

The resulting velocity of the ball after it hits the racket was of V= 51.6 m/s

Explanation:

m= 55.6 g = 0.0556 kg

t= 2.8 ms = 2.8 * 10⁻³ s

F= 1290 N/ms * t - 330 N/ms² * t²

F= 1024.8 N

F*t= m * V

V= F*t/m

V= 51.6 m/s

6 0

3 years ago

A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr

VLD [36.1K]

Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

$rR:\\\\E=k\frac{Q}{r^2}$

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

$F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}$

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

$Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}$

with this values of a you can use the following formula:

$a=\frac{v-v_o}{t}\\\\t=\frac{v-v_o}{a}=\frac{2550*10^3m/s-0m/s}{9.87*10^{11}m/s^2}=2.58*10^{-6}s$

hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s

3 0

3 years ago

In a real system of levers, wheels, or pulleys, the AMA is less than the IMA because _____.

pickupchik [31]

In a real system of levers, wheels, or pulleys, the AMA is less than the IMA because of friction.
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3 years ago

What is the difference between a low tide and a high tide

Savatey [412]

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3 years ago

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lina2011 [118]

Is this a multiple choice question? Or would you just like me to state generally the greatest influence over how big a baby elephant will grow ?

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3 years ago