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3 years ago

URGENT HELP NEEDED PLEASE!

Physics

1 answer:

GenaCL600 [577]3 years ago

8 0

Answer:

the total energy before collision is 12.25J

the total energy after collision is tranfered to sound and heat (by the friction between the two clay balls)

Explanation:

first we should know the velocity of the 2 balls:

momentum=mass ×velocity

we have the momentum and the mass of the 2

so momentum ÷mass=velocity

clay ball 1 velocity=7 m/s & clay ball 2 velocity= -7 m/s

KE=1/2 mv²

1/2 ×0.25×7²=6.125J KE of ball 1

1/2 ×0.25×-7²=6.125J KE of ball 2

add them both we will have 12.25J which is before collision

this energy will then be tranfered to heat and sound as they stop after the collision(velocity =0m/s)

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3 years ago

A 3.91 kg cart is moving at 5.7 m/s when it collides with a 4 kg cart which was at rest. They collide and stick together.

Nesterboy [21]

Answer:

<em>The velocity after the collision is 2.82 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

It states the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is

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If we have a system of two bodies, then the total momentum is the sum of the individual momentums:

$P=m_1v_1+m_2v_2$

If a collision occurs and the velocities change to v', the final momentum is:

$P'=m_1v'_1+m_2v'_2$

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2$

If both masses stick together after the collision at a common speed v', then:

$m_1v_1+m_2v_2=(m_1+m_2)v'$

The common velocity after this situation is:

$\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

There is an m1=3.91 kg car moving at v1=5.7 m/s that collides with an m2=4 kg cart that was at rest v2=0.

After the collision, both cars stick together. Let's compute the common speed after that:

$\displaystyle v'=\frac{3.91*5.7+4*0}{3.91+4}$

$\displaystyle v'=\frac{22.287}{7.91}$

$\boxed{v' = 2.82\ m/s}$

The velocity after the collision is 2.82 m/s

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3 years ago

Suppose that two point charges, each with a charge of +1.00 C, are separated by a distance of 1.0 m. If the distance between the

Aleksandr-060686 [28]

Given:

The magnitude of each charge is q1 = q2 = 1 C

The distance between them is r = 1 m

To find the force when distance is doubled.

Explanation:

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The force can be calculated by the formula

$F=k\frac{q1q2}{(r^{\prime})^2}$

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1 year ago

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Answer:

Explanation:

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3 years ago

Please help ASAP!

Softa [21]

Answer:

100 newton

Explanation:

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3 years ago

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