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ser-zykov [4K]
3 years ago
12

Three forces acting at a point keep it in equilibrium . If the angle between two of the forces , 3 N and 3 Nis 120°, then the th

ird force
is
Physics
1 answer:
Gennadij [26K]3 years ago
8 0

Answer:

  see attachments

Explanation:

The force necessary to maintain equilibrium depends on the directions of the two forces that are separated by 120°. If both are pointing toward or away from the point, the third force will be pointing that same direction. It will be 120° from the other two, and the same magnitude (3 N).

If the one of the two forces is pointing toward, and the other is pointing away, then the third force will have magnitude 3√3 N, and will be 30° from one of the other forces. This is illustrated in the 2nd attachment.

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two astronauts are taking a spacewalk outside the International Space Station the first astronaut has a mass of 64 kg the second
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Answer:

Approximately 0.88\; {\rm m \cdot s^{-1}} to the right (assuming that both astronauts were originally stationary.)

Explanation:

If an object of mass m is moving at a velocity of v, the momentum p of that object would be p = m\, v.

Since momentum of this system (of the astronauts) conserved:

\begin{aligned} &(\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\end{aligned}.

Assuming that both astronauts were originally stationary. The total initial momentum of the two astronauts would be 0 since the velocity of both astronauts was 0\!.

Therefore:

\begin{aligned} &(\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\\ &= 0\end{aligned}.

The final momentum of the first astronaut (m = 64\; {\rm kg}, v = 0.8\; {\rm m\cdot s^{-1}} to the left) would be p_{1} = m\, v = 64\; {\rm kg} \times 0.8\; {\rm m\cdot s^{-1}} = 51.2\; {\rm kg \cdot m \cdot s^{-1}} to the left.

Let p_{2} denote the momentum of the astronaut in question. The total final momentum of the two astronauts, combined, would be (p_{1} + p_{2}).

\begin{aligned} & p_{1} + p_{2} \\ &= (\text{Total Final Momentum}) \\ &= (\text{Total Initial Momentum})\\ &= 0\end{aligned}.

Hence, p_{2} = (-p_{1}). In other words, the final momentum of the astronaut in question is the opposite of that of the first astronaut. Since momentum is a vector quantity, the momentum of the two astronauts magnitude (51.2\; {\rm kg \cdot m \cdot s^{-1}}) but opposite in direction (to the right versus to the left.)

Rearrange the equation p = m\, v to obtain an expression for velocity in terms of momentum and mass: v = (p / m).

\begin{aligned}v &= \frac{p}{m} \\ &= \frac{51.2\; {\rm kg \cdot m \cdot s^{-1}}}{64\; {\rm kg}} && \genfrac{}{}{0}{}{(\text{to the right})}{} \\ &\approx 0.88\; {\rm m\cdot s^{-1}} && (\text{to the right})\end{aligned}.

Hence, the velocity of the astronaut in question (m = 58.2\; {\rm kg}) would be 0.88\; {\rm m \cdot s^{-1}} to the right.

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2 years ago
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