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2 years ago

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

Physics

2 answers:

vagabundo [1.1K]2 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]2 years ago

5 0

Answer:

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

Explanation:

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Consider a steel guitar string of initial length L=1.00 meter and cross-sectional area A=0.500 square millimeters. The Young's m

Reil [10]

Answer:

$\Delta l=0.015m$

Explanation:

We have given initial length of the steel guitar l = 1 m

Cross sectional area $A=0.5mm^2=0.5\times 10^{-6}m^2$

Young's modulus $\gamma=2\times 10^{11}Pa$

Force F = 1500 N

So stress $=\frac{force}{area}=\frac{1500}{0.5\times 10^{-6}}=3000\times 10^{-6}=3\times 10^{9}Pa$

We know that young's modulus $=\frac{stress}{strain}$

So $2\times 10^{11}=\frac{3\times 10^{9}}{strain}$

$strain=1.5\times 10^{-2}=0.015m$

Now strain $=\frac{\Delta l}{l}$

$0.015=\frac{\Delta l}{1}$

$\Delta l=0.015m$

6 0

3 years ago

Read 2 more answers

What factor affects the polarization of light by scattering?

MatroZZZ [7]

Answer:

The correct answer would be A.

Explanation:

I have gotten this answer many times and have never failed once just trust me.

4 0

1 year ago

Q7) A box sliding with a velocity of 5 m/s accelerates at 2 m/s^2. How

grigory [225]

Answer:

The box displacement after 6 seconds is 66 meters.

Explanation:

Let suppose that velocity given in statement represents the initial velocity of the box and, likewise, the box accelerates at constant rate. Then, the displacement of the object ( $\Delta s$ ), in meters, can be determined by the following expression:

$\Delta s = v_{o}\cdot t+\frac{1}{2}\cdot a\cdot t^{2}$ (1)

Where:

$v_{o}$ - Initial velocity, in meters per second.

$t$ - Time, in seconds.

$a$ - Acceleration, in meters per square second.

If we know that $v_{o} = 5\,\frac{m}{s}$ , $t = 6\,s$ and $a = 2\,\frac{m}{s^{2}}$ , then the box displacement after 6 seconds is:

$\Delta s = 66\,m$

The box displacement after 6 seconds is 66 meters.

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3 years ago

Inserting the formulas you found for Xman(t) and Xbus(t) into the conditionXman(tcatch)=Xbus(tcatch) , you obtain the following-

lara [203]

Answer:

c > √(2ab)

Explanation:

In this exercise we are asked to find the condition for c in such a way that the results have been real

The given equation is

½ a t² - c t + b = 0

we can see that this is a quadratic equation whose solution is

t = [c ±√(c² - 4 (½ a) b)] / 2

for the results to be real, the square root must be real, so the radicand must be greater than zero

c² -2a b > 0

c > √(2ab)

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2 years ago

What type of motion does the galaxy go through

Mariana [72]

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3 years ago