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3 years ago

Conservation of Momentum

Physics

1 answer:

aksik [14]3 years ago

3 0

Hi there!

This collision is an example of an inelastic collision since kinetic energy is lost from the collision.

We can represent this using the conservation of momentum formula:

m1v1 + m2v1 = m1vf + m2vf

Where:

m1 = blue ball

m2 = green ball

We know that the final velocity of the blue ball is 0, so:

m1v1 + m2v1 = m2vf

Rearrange to solve for the speed of the green ball:

(m1v1 + m2v1)/m2 = vf

Plug in given values:

((0.15 · 3) + (0.15 · 2)) / 0.15 = 5 m/s

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In order for work to be done, a force must cause movement true or false

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3 years ago

wo parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the magnitude

melisa1 [442]

Answer:

-18896.49 V/m

Explanation:

Distance between the two plates = 10 cm = 10 x $10^{-2}$ m = 0.1 m

Also, one of the plates is taken as zero volt.

a. The potential strength between the zero volt plate, and 7.05 cm (0.0705 m) away is 393 V

b. The potential strength between the other plate, and 2.95 cm (0.0295 m) away is 393 V

Potential field strength = -dV/dx

where dV is voltage difference between these points,

dx is the difference in distance between these points

For the first case above,

potential field strength = -393/0.0705 = -5574.46 V/m

For the second case ,

potential field strength = -393/0.0295 = -13322.03 V/m

Magnitude of the field strength across the plates will be

-5574.46 + (-13322.03) = -5574.46 + 13322.03 = -18896.49 V/m

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4 years ago

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3 years ago

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Read 2 more answers

If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa

shepuryov [24]

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

$g=G\frac{m}{r^{2}}$

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

$g_{E}=G\frac{m}{r_{E}^{2}}$

$g_{P}=G\frac{m}{r_{P}^{2}}$

The problem tells us the radius of the planet is twice that of the radius on earth, so:

$r_{P}=2r_{E}$

If we substituted that into the gravity of the planet equation we would end up with the following formula:

$g_{P}=G\frac{m}{(2r_{E})^{2}}$

Which yields:

$g_{P}=G\frac{m}{4r_{E}^{2}}$

So we can now compare the two gravities:

$\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}$

When simplifying the ratio we end up with:

$\frac{g_{P}}{g_{E}}=\frac{1}{4}$

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

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3 years ago