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3 years ago

8

A man inside the water seea a fish in water. will the fish appear to man in it's actual position? will the man appear to fish in

his actual position?

2 answers:

weqwewe [10]3 years ago

4 0

Answer:

no

Explanation:

due to the refraction of light,meaning bending of light when it passes from diffrent mediums with diffrent densities

(probably mot the complete ans but smtg like this)

fiasKO [112]3 years ago

3 0

No man will not even go near and fish will but not same position

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Read 2 more answers

In an elastic collision, a 300 kg bumper car collides directly from behind with a second, identical bumper car that is traveling

evablogger [386]

Answer:

If we had:

$v_{1i}=5.3m/s$

$v_{2i}=5.9m/s$

We will have:

$v_{1f}=5.9m/s$

$v_{2f}=5.3m/s$

Explanation:

In an elastic collision both linear momentum and kinetic energy are conserved, so we will have:

$p_i=p_f$

$K_i=K_f$

We will call our bumpers 1 and 2.

For the momentum equation we know that:

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$

Since all the masses are the same (300kg), they cancel out:

$v_{1i}+v_{2i}=v_{1f}+v_{2f}$

For the kinetic energy equation we know that:

$\frac{m_1v_{1i}^2}{2}+\frac{m_2v_{2i}^2}{2}=\frac{m_1v_{1f}^2}{2}+\frac{m_2v_{2f}^2}{2}$

Since all the masses are the same (300kg), they cancel out (and also the 2 dividing):

$v_{1i}^2+v_{2i}^2=v_{1f}^2+v_{2f}^2$

We then must solve this system:

$v_{1i}+v_{2i}=v_{1f}+v_{2f}$

$v_{1i}^2+v_{2i}^2=v_{1f}^2+v_{2f}^2$

Which we will rewrite as:

$v_{1i}-v_{1f}=v_{2f}-v_{2i}$

$v_{1i}^2-v_{1f}^2=v_{2f}^2-v_{2i}^2$

The last of these equations can be written as:

$(v_{1i}+v_{1f})(v_{1i}-v_{1f})=(v_{2f}+v_{2i})(v_{2f}-v_{2i})$

But we know that $v_{1i}-v_{1f}=v_{2f}-v_{2i}$ , so those cancel out:

$v_{1i}+v_{1f}=v_{2f}+v_{2i}$

So we can write:

$v_{1i}-v_{1f}+v_{2i}=v_{2f}$

$v_{1i}+v_{1f}-v_{2i}=v_{2f}$

Which means:

$v_{1i}-v_{1f}+v_{2i}=v_{1i}+v_{1f}-v_{2i}$

Which solving for the final velocity leaves us with:

$v_{2i}+v_{2i}=+v_{1f}+v_{1f}$

$v_{1f}=v_{2i}$

Grabbing any equation that relates both final velocities easily, for example $v_{1i}-v_{1f}+v_{2i}=v_{2f}$ , we obtain:

$v_{2f}=v_{1i}-v_{1f}+v_{2i}=v_{1i}-v_{1f}+v_{1f}=v_{1i}$

So we conclude that the bumpers have just exchanged velocities (something sometimes seen in billiards for example):

$v_{1f}=v_{2i}=5.9m/s$

$v_{2f}=v_{1i}=5.3m/s$

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