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2 years ago

Help ASAP Answer question 3 please!!!

Physics

1 answer:

PolarNik [594]2 years ago

8 0

Answer:

the answer is ture

Explanation:

because the energy substances must absorb in order to change from liquid to gas

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What two elements are named after Dimitri Mendeleev?

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I'm pretty sure there is only one element named after Mendeleev: <span>Mendelevium.</span>

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3 years ago

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a steel ball bearing is released from a height hhh and rebounds after hitting a steel plate to a height hhh. what is true about

lbvjy [14]

In collision of the steel ball and the steel plate, the collision is an inelastic collision and there is loss in the kinetic energy.

<h3>What are collisions?</h3>

Collisions occur when two objects that are moving in the same directions or in different direction meet each other and collide.

There are two types of collisions:

elastic collision - the kinetic energy is conserved
inelastic collision - there is a loss in kinetic energy

In the collision of the steel ball and the steel plate, there is loss in the kinetic energy of the steel ball which is converted to sound energy.

In conclusion, the collision of the steel and steel plate is an inelastic collision.

Learn more about collisions at: brainly.com/question/7694106

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1 year ago

A hot-air balloon is rising upward with a constant speed of 2.03 m/s. When the balloon is 8.13 m above the ground, the balloonis

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2 years ago

What is the pressure 100m below the surface of the sea , if the density of sea water is 1150kg/m3

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2 years ago

Block A has a mass of 0.5kg, and block B has a mass of 2kg. Block is is released at a height of 0.75 meters above B. The coeffic

VikaD [51]

Answer:

0.075 m

Explanation:

The picture of the problem is missing: find it in attachment.

At first, block A is released at a distance of

h = 0.75 m

above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:

$m_Agh=\frac{1}{2}m_Av_A^2$

where

$g=9.8 m/s^2$ is the acceleration due to gravity

$m_A=0.5 kg$ is the mass of the block

$v_A$ is the speed of the block A just before touching block B

Solving for the speed,

$v_A=\sqrt{2gh}=\sqrt{2(9.8)(0.75)}=3.83 m/s$

Then, block A collides with block B. The coefficient of restitution in the collision is given by:

$e=\frac{v'_B-v'_A}{v_A-v_B}$

where:

e = 0.7 is the coefficient of restitution in this case

$v_B'$ is the final velocity of block B

$v_A'$ is the final velocity of block A

$v_A=3.83 m/s$

$v_B=0$ is the initial velocity of block B

Solving,

$v_B'-v_A'=e(v_A-v_B)=0.7(3.83)=2.68 m/s$

Re-arranging it,

$v_A'=v_B'-2.68$ (1)

Also, the total momentum must be conserved, so we can write:

$m_A v_A + m_B v_B = m_A v'_A + m_B v'_B$

where

$m_B=2 kg$

And substituting (1) and all the other values,

$m_A v_A = m_A (v_B'-2.68) + m_B v_B'\\v_B' = \frac{m_A v_A +2.68 m_A}{m_A + m_B}=1.30 m/s$

This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to

$\frac{1}{2}m_B v_B'^2 = \frac{1}{2}kx^2$

where

k = 600 N/m is the spring constant

x is the compression of the spring

And solving for x,

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(2)(1.30)^2}{600}}=0.075 m$

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2 years ago