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kvasek [131]
3 years ago
15

A stone on ground has zero energy​

Physics
1 answer:
Len [333]3 years ago
5 0

Answer:

Since energy means capacity to do work, when stone is in ground level, it has 0 work because it doesn't covered any distance but when stone lifted at a certain height , it covered a certain distance and work is done. Thus same stone has energy when it is lifted at a height .

Explanation:

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Define the Element of Art, value. If you were using a pencil, how would you create a darker value?
ankoles [38]

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You would create a darker value if you shaded it, shading it makes it darker.

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3 years ago
In space (no gravity or friction), you throw a ball with mass 0.1 kg at a target with mass 1 kg. You throw the ball at a speed o
ioda

Answer:

Explanation:

We shall apply law of conservation of  momentum in space to know the velocity of combination after the impact

m₁v₁ = m₂v₂

.1 x 4 = ( 1 + .1 ) v₂

v₂ = .3636 m /s

1  )  

Kinetic energy of the combination

= 1/2 x 1.1 x ( .3636)²

= 7.3 x 10⁻² J

2 )

Initial kinetic energy of the system

= 1/2 x 0.1 x 4²

= 0.8 J

Final  kinetic energy of the system = 7.3 x 10⁻²

Loss of energy = .8 - .073

= .727 J

This energy was converted into internal energy of the system .

3 )

increase in entropy = dQ / T

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6 0
4 years ago
A disc of mass m slides with negligible friction along a flat surface with a velocity v. The disc strikes a wall head-on and bou
qwelly [4]

Answer:

-v/2

Explanation:

Given that:

  • a disc of mass m
  • Collides with the wall going through a sliding motion on on the plane smooth surface.
  • Upon rebounding from the wall its kinetic energy becomes one-fourth of the initial kinetic energy before collision.

<u>We know, kinetic energy is given as:</u>

KE_i=\frac{1}{2}. m.v^2

consider this to be the initial kinetic energy of the body.

<u>Now after collision:</u>

KE_f=\frac{1}{4}\times KE_i

KE_f=\frac{1}{4} \times \frac{1}{2}\times m.v^2

Considering that the mass of the body remains constant before and after collision.

KE_f=\frac{1}{2}\times m.(\frac{v}{2})^2

Therefore the velocity of the body after collision will become half of the initial velocity but its direction is also reversed which can be denoted by a negative sign.

3 0
3 years ago
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