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Ksenya-84 [330]
3 years ago
13

A block with a mass of 8.7 kg is dropped from rest from a height of 8.7 m, and remains at rest after hitting the ground. 1)If we

consider the system to consist of the block, the ground, and the surrounding air, what is the change in the internal energy of the system
Physics
1 answer:
Harlamova29_29 [7]3 years ago
7 0

To solve this problem we will apply the concepts related to gravitational potential energy.

This can be defined as the product between mass, gravity and body height.

Mathematically it can be expressed as

\Delta P = mgh

\Delta P = (8.7)(9.8)(3)

\Delta P = 255.78J

Therefore the change in the internal energy of the system is 255.78

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telo118 [61]
<span>Avogadro's number represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. This number can be used to convert the number of atoms or molecules into number of moles.

 </span>1.83 moles K (6.022 x 10^23 atoms / mole ) = 1.10 x 10^24 atoms K
4 0
3 years ago
The tow spring on a car has a spring constant of 3,086 N / m and is initially stretched 18.00 cm by a 100.0 kg college student o
sdas [7]

Answer:

The velocity of the skateboard is 0.774 m/s.

Explanation:

Given that,

The spring constant of the spring, k = 3086 N/m

The spring is stretched 18 cm or 0.18 m

Mass of the student, m = 100 kg

Potential energy of the spring, P_f=20\ J

To find,

The velocity of the car.

Solution,

It is a case of conservation of energy. The total energy of the system remains conserved. So,

P_i=K_f+P_f

\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2+20

\dfrac{1}{2}\times 3086\times (0.18)^2=\dfrac{1}{2}mv^2+20

50-20=\dfrac{1}{2}mv^2

30=\dfrac{1}{2}mv^2

v=\sqrt{\dfrac{60}{100}}

v = 0.774 m/s

So, the velocity of the skateboard is 0.774 m/s.

7 0
3 years ago
A 0.14-MIN baseball is dropped from rest. It has a momentum of 0.90 kg⋅m/skg⋅m/s just before it lands on the ground.
nikitadnepr [17]

The time spent in the air by the ball at the given momentum is 6.43 s.

The given parameters;

  • <em>momentum of the ball, P = 0.9 kgm/s</em>
  • <em>weight of the ball, W = 0.14 N</em>

The impulse experienced by the ball is calculated as follows;

Ft = \Delta P

where;

Ft is impulse

\Delta P is change in momentum

The time of motion of the ball is calculated as follows;

t = \frac{\Delta P}{F} \\\\t = \frac{0.9 - 0}{0.14} \\\\t = 6.43 \ s

Thus, the time spent in the air by the ball at the given momentum is 6.43 s.

Learn more here:brainly.com/question/13468390

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The point at which all motion stops.
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The Sun is to a planet as -
serg [7]

Answer:a

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