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ankoles [38]
3 years ago
14

URGENT!! An astronaut on the International Space Station is doing a spacewalk to fix a solar panel that has malfunctioned. While

he is completing his task, a piece of space junk flies past and cuts the line that has him tethered to the space station. He begins to float away and must think quickly in order to get himself back to safety. All the astronaut has with him on his suit are some small tools and a can of compressed air.
What would be the most effective way for the astronaut to get himself back to safety?
A. use a swimming motion to move himself back to the space station
b. toss his tools in the direction of the space station
c. shoot gas from his compressed air canister towards the space station
d. shoot gas from his compressed air canister away from the space station
Physics
2 answers:
Eddi Din [679]3 years ago
5 0

Answer:

D. Shoot gas from his compressed air canister away from the space station

Explanation:

Isacc Newton's 3rd law states "For every action, there is an equal and opposite reaction." The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object.

aleksley [76]3 years ago
3 0

C because the air will push him closer to the space station

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What is the electric potential energy of an electron at the negative end of the cable, relative to the positive end of the cable
VashaNatasha [74]

Answer:

Electric potential energy at the negative terminal: 1.92\cdot 10^{-18}J

Explanation:

When a particle with charge q travels across a potential difference \Delta V, then its change in electric potential energy is

\Delta U = q \Delta V

In this problem, we know that:

The particle is an electron, so its charge is

q=-1.60\cdot 10^{-19}C

We also know that the positive terminal is at potential

V_+=0V

While the negative terminal is at potential

V_-=-12 V

Therefore, the potential difference (final minus initial) is

\Delta V = -12-0 = -12 V

So, the change in potential energy of the electron is

\Delta U = (-1.6\cdot 10^{-19})(-12)=1.92\cdot 10^{-18}J

This means that the electron when it is at the negative terminal has 1.92\cdot 10^{-18}J of energy more than when it is at the positive terminal.

Since the potential at the positive terminal is 0, this means that the electric potential energy of the electron at the negative end is

1.92\cdot 10^{-18}J

3 0
3 years ago
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Vilka [71]
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3 years ago
Masses A and B rest on very light pistons that enclose a fluid.There is no friction between the pistons and the cylinders they f
RSB [31]

Answer:

D)Not enough information

Explanation:

According to Pascal's principle, the pressure exerted on the two pistons is equal:

p_A = p_B

Pressure is given by the ratio between force F and area A, so we can write

\frac{F_A}{A_A}=\frac{F_B}{A_B}

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We also know that A_B = 2.0 m^2\\A_A = 1.0 m^2

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3 years ago
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