The vertical velocity

A billiard ball collides with a second identical ball in an elastic head-on collision. What is the kinetic energy of the system

murzikaleks [220]

Answer:

Explanation:

From the question, we were made to understand that the collision between the two billiard balls was an elastic collision. Hence, an elastic collision is one in which the kinetic energy is conserved. Meaning the kinetic energy before the collision is still retained after the collision.

Kinetic energy before collision = kinetic energy after collision

1/2mv^2 = 1/2mv^2

There was no gain nor loss in energy

5 0

3 years ago

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If the spring constant is doubled, what value does the period have for a mass on a spring? A. The period would double by square

antoniya [11.8K]

Answer:

The period would decrease by sqrt(2)

Explanation:

The restoring force is given by,

F = -kx

According to Newton's second law of motion,

ma = -kx

ma + kx = 0

The time period is given by,

T = $\frac{2\pi }{\omega }$

Where $\omega$ is the angular velocity and it is given by,

$\omega$ = $\sqrt{\frac{k}{m} }$

Now if the spring constant is doubled then,

$k_{2} = 2k$

Thus,

$T_{2}$ = $\frac{2\pi }{\sqrt{\frac{2k}{m} } }$

$\frac{T_{2} }{T} = \frac{\frac{2\pi }{\sqrt{\frac{2k}{m} } }}{\frac{2\pi }{\sqrt{\frac{k}{m} } }}$

$\frac{T_{2} }{T} = \sqrt{\frac{k}{2k} } = \sqrt{\frac{1}{2} }$

$T_{2} = \frac{T}{\sqrt{2} }$

Thus, The period would decrease by sqrt(2).

Hence, option D is correct.

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3 years ago

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Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.

8_murik_8 [283]

The electric potential is a scalar unit, so we don't have to struggle with the vectors. The formula that gives electric potential is

$V = \frac{1}{4\pi\epsilon_0}\frac{q}{r}$

1) At point a, the electric potential is the sum of the potentials due to q1 and q2. So,

$V_a = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}$

The distance from the center of the square to one of the corners is $\sqrt2 L/2 = 0.035m$

$V_a = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.035} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.035} = 0$

The answer is zero, because the point charges are at equal distances and their magnitudes are also equal but their directions are opposite.

2) $V_b = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}$

$r_1 = 0.05\sqrt2m\\r_2 = 0.05m$

$V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05\sqrt2} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.05}\\V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05} (\frac{1}{\sqrt2}-1)\\V_b = \frac{1}{4\pi\epsilon_0} (4\times 10^{-5})(-0.29)\\V_b = (-\frac{2.9\times10^{-6}}{\pi\epsilon_0})[tex]3) The work done on q3 by q1 and q2 is equal to the difference between energies. This is the work-energy theorem. So,[tex]W = U_b - U_a$

$U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_3}{r} = Vq_3$

$W = q_3(V_b - V_a) = q_3(V_b - 0)\\W = (-2\times10^{-6})(-\frac{2.9\times10^{-6}}{\pi\epsilon_0})\\W = \frac{5.8\times10^{-12}}{\pi\epsilon_0}$

4 0

3 years ago

How is energy conserved in a transformation?

irina [24]

As the water plunges, its velocity increases. Its potential energy becomes kineticenergy. The law of conservation of energy states that when one form of energy istransformed to another, no energy is destroyed in the process. ... So the total amount of energy is the same before and after any transformation.

hope it helps

5 0

3 years ago

Humorous name for a model T Ford

AfilCa [17]

Old Grandpy!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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3 years ago

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