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4 years ago

1- a 0.250 kg egg-drop project is at the top of the 14 meter tall stadium. what is the projects potential energy?

1 answer:

8 0

<span>1. The potential energy is 34.3J

2. The kinetic energy is 61,875J.

3. A 0.5 kg ball is at the top of a 20 meter tall hill.

a) What is the ball's kinetic energy ? The ball's kinetic energy is zero.
b) what is the ball's potential energy ? The ball's potential energy is 98.1J.
c) what is the ball's mechanical energy ? The ball's mechanical energy is 98.1J.

4. 2.0 kg toy has been thrown in the air it is 1.5m above the ground and is moving up at 3.0 m/s

a) What is the toy's kinetic energy ? The toy's kinetic energy is 9J.
b) what is the toy's potential energy ? The toy's potential energy is 29.43J.
c) what is the toy's mechanical energy ? The toy's mechanical energy s 38.43J.</span>

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Two pieces of clay are moving directly toward each other. When they collide, they stick together and move as one piece. One piec

Wittaler [7]

Answer:

The fraction of the total initial kinetic energy is lost during the collision is $\dfrac{11}{17}\ J$

Explanation:

Given that,

Mass of one piece = 300 g

Speed of one piece = 1 m/s

Mass of other piece = 600 g

Speed of other piece = 0.75 m/s

We need to calculate the final velocity

Using conservation of energy

$m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v$

Put the value intro the formula

$300\times10^{-3}\times1+600\times10^{-3}\times(0.75)=(300\times10^{-3}+600\times10^{-3})v$

$v=\dfrac{00\times10^{-3}\times1+600\times10^{-3}\times(-0.75)}{(300\times10^{-3}+600\times10^{-3})}$

$v=-0.5\ m/s$

We need to calculate the total initial kinetic energy

Using formula of kinetic energy

$K.E_{i}=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2$

Put the value into the formula

$K.E_{i}=\dfrac{1}{2}\times300\times10^{-3}\times1^2+\dfrac{1}{2}\times600\times10^{-3}\times(0.75)^2$

$K.E_{i}=0.31875\ J$

We need to calculate the total final kinetic energy

Using formula of kinetic energy

$K.E_{f}=\dfrac{1}{2}(m_{1}+m_{2})v^2$

Put the value into the formula

$K.E_{f}=\dfrac{1}{2}\times(300\times10^{-3}+600\times10^{-3})\times(-0.5)^2$

$K.E_{f}=0.1125\ J$

We need to calculate the energy lost during the collision

Using formula of energy lost

$energy\ lost=\dfrac{0.31875-0.1125}{0.31875}$

$energy\ lost=\dfrac{11}{17}\ J$

Hence, The fraction of the total initial kinetic energy is lost during the collision is $\dfrac{11}{17}\ J$

3 0

3 years ago

What is the change in internal energy if 30 J of thermal energy is released

alexira [117]

Answer:

ΔU = -70 J

Explanation:

ΔU = Q − W

where ΔU is the change in internal energy,

Q is the heat absorbed by the system,

and W is the work done by the system (on the surroundings).

30 J of thermal energy is released, so Q = -30 J.

40 J of work is done by the system, so W = 40 J.

Therefore, the change in internal energy is:

ΔU = -30 J − 40 J

ΔU = -70 J

6 0

3 years ago

According to Boyles’ law, PV = constant. If a graph is plotted with the pressure P against the volume V, the graph would be a(n)

Hunter-Best [27]

Answer:

C. hyperbola

Explanation:

From Boyle's law:

PV = k, where k is a constant

Solving for P:

P = k / V

At first glance, this equation doesn't fit any of the options. But when you graph it, you can see that it's actually a <em>rotated</em> hyperbola.

4 0

3 years ago

Read 2 more answers

Calculate the electric field on the surface of a charged sphere with a radius of 5cm and charge 10μC. Thanks

victus00 [196]

As you approach the surface of the sphere very closely, the electric field should resemble more and more the electric field from an infinite plane of charge.

If you check Gauss's law (recalling that the field in the conductor is zero) you will see that if the surface charge density is σ=Q/4πR2, then indeed the field at the surface is σ/ϵ0 as in the infinite charge of plane case.

Such a field is constant, the field lines are parallel and non-diverging, and the infinities associated with the field due to point charge do not arise.

Explanation:

4 0

2 years ago

A 0.9 kg metal bolt is heated to an unknown initial temperature. It is then dropped into a Calorimeter containing 0.15 kg of wat

Montano1993 [528]

Answer:

Whets nasılsın merhaba arkadaşlar ben bu etikete yazar nasılsın

8 0

3 years ago