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3 years ago

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The total momentum of two marbles before a collision is 0.06 kg-m / s. no outside forces act on the marbles. what is the total m

omentum of the marbles after the collision?\

1 answer:

SpyIntel [72]3 years ago

4 0

Hi my friend, since momentum is always conserved without external forces, the momentum after the collosion will still be 0.06 kg*m/s. Hope it helps☺

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How do icebergs form

pishuonlain [190]

Answer:

They break off large ice sheets found at the North and South poles.

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2 years ago

A less massive moving object has an elastic collision with a more massive object that is not moving. Compare the initial velocit

stepladder [879]

Assume that the small-massed particle is $m$ and the heavier mass particle is $M$ .

Now, by momentum conservation and energy conservation:

$mv = mv_{m} + Mv_{M}$

$mv^{2} = mv^{2}_{m} + Mv^{2}_{M}$

Now, there are 2 solutions but, one of them is useless to this question's main point so I excluded that point. Ask me in the comments if you want the excluded solution too.

$v_{m} = v\frac{m - M}{m + M}\\\\\\v_{M} = v\frac{m + M}{m + M}\\$

So now, we see that $v_{m} < 0$ and $v_{M} > 0$ . So therefore, the smaller mass recoils out.

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3 years ago

A car accelerates from 13 m/s to 25 m/s in 5.0 s. assume constant acceleration. what was its acceleration?

natima [27]

<span>a = 25-13/6 = 12/6 = 2 m/s^2
Av speed: 25+13/2 = 38/2 = 19 m/sec
Dist = speed * time
19 * 6 = 114 meters</span>

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3 years ago

To calibrate your calorimeter cup, you first put 45 mL of cold water in the cup, and measure its temperature to be 24.7 °C. You

drek231 [11]

Answer : The heat change of the cold water in Joules is, $1.6\times 10^3J$

Explanation :

First we have to calculate the mass of cold water.

As we know that the density of water is 1 g/mL. The volume of cold water is 45 mL.

$Density=\frac{Mass}{Volume}$

$Mass=Density\times Volume=1g/mL\times 45mL=45g$

Now we have to calculate the heat change of cold water.

Formula used :

$Q=m\times c\times (T_2-T_1)$

where,

Q = heat change of cold water = ?

m = mass of cold water = 45 g

c = specific heat of water = $4.184J/g^oC$

$T_1$ = initial temperature of cold water = $24.7^oC$

$T_2$ = final temperature = $33.4^oC$

Now put all the given value in the above formula, we get:

$Q=45g\times 4.184J/g^oC\times (33.4-24.7)^oC$

$Q=1638.036J=1.6\times 10^3J$

Therefore, the heat change of cold water is $1.6\times 10^3J$

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3 years ago

Which statement about subatomic particles is true?

navik [9.2K]

A. An electron has far less mass than either a proton or neutron.

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3 years ago