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Yuri [45]
3 years ago
14

Why are astronauts weightless in the Space Station? You may find it helpful to watch the video "Newton's Laws of Motion." Why ar

e astronauts weightless in the Space Station? You may find it helpful to watch the video "Newton's Laws of Motion."
A) Because the Space Station is moving at constant velocity.
B) Because the Space Station is constantly in free-fall around the Earth.
C) Because there is no gravity in space.
D) Because the Space Station is traveling so fast.
Physics
1 answer:
Zolol [24]3 years ago
7 0

Answer:

B) Because the Space Station is constantly in free-fall around the Earth.

Explanation:

Anything that is falling experiences an upward force on them. For example when a person is going down in a lift they will experience something that is pushing them upwards. This happens due to the fact that the total acceleration the body is feeling is less than the acceleration due to graviity.

The force on a body which is falling is

F=m(g-a)

Where,

m = Mass of object

g = acceleration due to gravity

a = acceleration the object is experiencing.

a = g. So, the force becomes zero and the object experiences weightlessness.

Hence, the astronauts in the space station experience weightlessness due to fact that the Space Station is constantly in free-fall around the Earth.

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by answering your question

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While visiting the beach, you enjoy the warm ocean water, but the sand burns your feet. That night you walk along the beach and
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Cindy saw her mother heating a wet pan on the stove. As the pan got hotter, the water on the outside began to dry. Why?
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Answer:

As the temperature of materials increase, the objects find a phenomenon called change of phase.

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3 years ago
Two protons are maintained at a separation of nm. Calculate the electric potential due to the two particles at the midpoint betw
Liono4ka [1.6K]

Answer:

The electric potential is approximately 5.8 V

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero

Explanation:

The two protons can be considered as point charges. Therefore, the electric potential is given by the point charge potential:

\displaystyle{U=\frac{q}{4\pi \epsilon_0r}} (1)

where q is the charge of the particle, \epsilon_0 the electric permittivity of the vacuum (I assuming the two protons are in a vacuum) and r is the distance from the point charge to the point where the potential is being measured. Because the electric potential is an scalar, we can simply add the contribution of the two potentials in the midpoint between the protons. Thus:

\displaystyle{U_{midpoint}=\frac{q}{4\pi \epsilon_0r}}+\frac{q}{4\pi \epsilon_0r}}=\frac{q}{2\pi \epsilon_0r}}}

Substituting the values q=1.602 \cdot10^{-19}\ C, \displaystyle{\frac{1}{4\pi\epsilon_0}=8.99\cdot 10^9 N\cdot m^2\cdot C^{-2}} and r=0.5 \cdot 10^{-9} m we obtain:

\displaystyle{U_{midpoint}=\frac{q}{2\pi \epsilon_0r}}=5.759 \approx 5.8 V}

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero.

6 0
3 years ago
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