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3 years ago

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A bumper has a non negligible mass. After a collision with an ideal spring, the spring is compressed to a distance of 90% of the

value of its distance when the bumper's mass was negligible. Give a reasonable explanation for this decrease.

1 answer:

Y_Kistochka [10]3 years ago

3 0

Deacceleration causing reverse of speed impact on the body of the frame causing reverse of shockwave.

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Explain the relation between Joule and Erg.

LekaFEV [45]

Answer:

Both erg and Joule are units of Energy in the CGS and S.I system respectively.

Joule and erg can also be written as (kg m2/s2) and (g cm2/s2) respectively.

Explanation:

1 Joule = 107 erg ⇒ 107 kg m2/s2.

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1 year ago

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HELP ASAP HELP Two tennis balls of the same mass are served at different speeds: 30 m/s and 60 m/s. Which serve has more kinetic

inna [77]

Answer:

<em>The second ball has four times as much kinetic energy as the first ball.</em>

Explanation:

<u>Kinetic Energy </u>

Is the type of energy an object has due to its state of motion. It's proportional to the square of the speed.

The equation for the kinetic energy is:

$\displaystyle K=\frac{1}{2}mv^2$

Where:

m = mass of the object

v = speed at which the object moves

The kinetic energy is expressed in Joules (J)

Two tennis balls have the same mass m and are served at speeds v1=30 m/s and v2=60 m/s.

The kinetic energy of the first ball is:

$\displaystyle K_1=\frac{1}{2}m\cdot 30^2$

$\displaystyle K_1=\frac{1}{2}m\cdot 900$

$K_1=450m$

The kinetic energy of the second ball is:

$\displaystyle K_2=\frac{1}{2}m\cdot 60^2$

$\displaystyle K_2=\frac{1}{2}m\cdot 3600$

$K_2=1800m$

Being m the same for both balls, the second ball has more kinetic energy than the first ball.

To find out how much, we find the ratio:

$\displaystyle \frac{K_2}{K_1}=\frac{1800m}{450m}$

Simplifying:

$\displaystyle \frac{K_2}{K_1}=4$

The second ball has four times as much kinetic energy as the first ball.

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3 years ago

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A 2.7-kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spri

exis [7]

a) The speed of the block at a height of 0.25 m is 2.38 m/s

b) The compression of the spring is 0.25 m

c) The final height of the block is 0.54 m

Explanation:

a)

We can solve the problem by using the law of conservation of energy. In fact, the total mechanical energy (sum of kinetic+gravitational potential energy) must be conserved in absence of friction. So we can write:

$U_i +K_i = U_f + K_f$

where

$U_i$ is the initial potential energy, at the top

$K_i$ is the initial kinetic energy, at the top

$U_f$ is the final potential energy, at halfway

$K_f$ is the final kinetic energy, at halfway

The equation can be rewritten as

$mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2$

where:

m = 2.7 kg is the mass of the block

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

$h_i = 0.54$ is the initial height

u = 0 is the initial speed

$h_f = 0.25 m$ is the final height of the block

v is the final speed when the block is at a height of 0.25 m

Solving for v,

$v=\sqrt{u^2+2g(h_i-h_f)}=\sqrt{0+2(9.8)(0.54-0.25)}=2.38 m/s$

b)

The total mechanical energy of the block can be calculated from the initial conditions, and it is

$E=K_i + U_i = 0 + mgh_i = (2.7)(9.8)(0.54)=14.3 J$

At the bottom of the ramp, the gravitational potential energy has become zero (because the final heigth is zero), and all the energy has been converted into kinetic energy. However, then the block compresses the spring, and the maximum compression of the spring occurs when the block stops: at that moment, all the energy of the block has been converted into elastic potential energy of the spring. So we can write

$E=E_e = \frac{1}{2}kx^2$

where

k = 453 N/m is the spring constant

x is the compression of the spring

And solving for x, we find

$x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(14.3)}{453}}=0.25 m$

c)

If there is no friction acting on the block, we can apply again the law of conservation of energy. This time, the initial energy is the elastic potential energy stored in the spring:

$E=E_e = 14.3 J$

while the final energy is the energy at the point of maximum height, where all the energy has been converted into gravitational potetial energy:

$E=U_f = mg h_f$

where $h_f$ is the maximum height reached. Solving for this quantity, we find

$h_f = \frac{E}{mg}=\frac{14.3}{(2.7)(9.8)}=0.54 m$

which is the initial height: this is correct, because the total mechanical energy is conserved, so the block must return to its initial position.

Learn more about kinetic and potential energy:

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3 years ago

Which device is based on the expansion of matter as temperature increases?

Nat2105 [25]

Thaattttttt would be mass vs volume im in science now. hope it helps

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3 years ago

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In a level parking lot, a man with a mass of 100 kg gives a shove to a boy with a mass of 50 kg who is on roller skates. The boy

natita [175]

Answer:

100N

Explanation:

Newton's third law states that whenever an object exerts a force on a second object, it exerts a force of equal magnitude and direction but in the opposite direction on the first. It is often stated as follows: Each action always opposes an equal but opposite reaction.

The subject 1 of 100kg is making a force F, to move an object from 50Kg to 2m / s ^ 2. This Force the object of 50Kg will reflect it in the opposite direction by Newton's third law.

Once the parameter of the force that both are experiencing is clarified, Newton's second law is applied to their respective calculation.

$F = ma = 50kg * 2m / s ^ 2 = 100N$

That is the force the boy exert on the man during the shove.

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3 years ago