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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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2 years ago
How do I do this physics problem about potential energy and kinetic energy?
larisa86 [58]

Ok i apologise for the messy working but I'll try and explain my attempt at logic

Also note i ignore any air resistance for this.

First i wrote the two equations I'd most likely need for this situation, the kinetic energy equation and the potential energy equation.

Because the energy right at the top of the swing motion is equal to the energy right in the "bottom" of the swing's motion (due to conservation of energy), i made the kinetic energy equal to the potential energy as indicated by Ek = Ep.

I also noted the "initial" and "final" height of the swing with hi and hf respectively.

So initially looking at this i thought, what the heck, there's no mass. Then i figured that using the conservation of energy law i could take the mass value from the Ek equation and use it in the Ep equation. So what i did was take the Ek equation and rearranged it for m as you can hopefully see. Then i substituted the rearranged Ek equation into the Ep equation.

So then the equation reads something like Ep = (rearranged Ek equation for m) × g (which is -9.81) × change in height (hf - hi).

Then i simplify the equation a little. When i multiply both sides by v^2 i can clearly see that there is one E on each side (at that stage i don't need to clarify which type of energy it is because Ek = Ep so they're just the same anyway). So i just canceled them out and square rooted both sides.

The answer i got was that the max velocity would be 4.85m/s 3sf, assuming no losses (eg energy lost to friction).

I do hope I'm right and i suppose it's better than a blank piece of paper good luck my dude xx

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