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TEA [102]
3 years ago
9

An elephant and a mouse would both have zero weight in gravity-free space. If they were moving toward you with the same speed, w

ould they bump into you with the same effect? Explain.
Physics
1 answer:
Dovator [93]3 years ago
8 0

The elephant and the mouse having zero weight in a gravity free space will not bump into you at the same effect.

<u>Explanation: </u>

When both are in a gravity free space, the weights are zero, as we know that the\text {weight of the body}=\text {mass of the body} \times \text {acceleration due to gravity}

\text {here, the weight of elephant}=\text {mass of elephant } \times \text {zero gravti} y=zero

\text {similarly,weight of mouse}=\text {mass of mouse } \times \text {zero gravity}=zero

But when they will acquire the speed of same magnitude, say v, their different masses will acquire different momentum, which will make the difference in effect while bumping.  

\text { momentum of elephant }=\text { mass of elephant } \times v  \text { momentum of mouse = mass of mouse } \times v

And as we know \text { mass of elephant }>\text { mass of mouse }  Therefore, effect of impact by elephant will be more than that of mouse . An elephant breaking into you will take you back faster than a mouse in space hits you.

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cestrela7 [59]

Frequency division multiplexing operates by dividing the signal into different frequencies

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The technique that is used in the networking is the Frequency Division Multiplexing. using this technique, the existing bandwidths can be partitioned into different frequency bandwidths. These are not interrupting with each other. Each bandwidth can be used for carrying signals individually.

Using this technique many users can share a particular communication medium and they will not be interrupted with each other's communication.Hence this technique can also be termed as Frequency Division Multiple Access.

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3 years ago
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3 years ago
A tube is sealed at both ends and contains a 0.0100-m long portion of liquid. The length of the tube is large compared to 0.0100
Ahat [919]

Answer:

31.321 rad/s

Explanation:

L = Tube length

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F=\dfrac{mv^2}{L}\\ F=\dfrac{m(\omega L)^2}{L}\\ F=m\omega^2\\ F= 0.01A\rho\omega^2L

Pressure is given by

P=\dfrac{F}{A}=\rho gL\\\Rightarrow \dfrac{0.01A\rho\omega^2L}{A}=\rho gL\\\Rightarrow 0.01\omega^2=g\\\Rightarrow \omega^2=\dfrac{g}{0.01}\\\Rightarrow \omega=\sqrt{\dfrac{g}{0.01}}\\\Rightarrow \omega=\sqrt{\dfrac{9.81}{0.01}}\\\Rightarrow \omega=31.321\ rad/s

The angular speed of the tube is 31.321 rad/s

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Give 10 examples of units you might use or see in any given day
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