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jasenka [17]
3 years ago
10

Pilings are driven into the ground at a building site by dropping a 2250 kg object onto them. What change in gravitational poten

tial energy does the object undergo if it is released from rest 16.0 m above the ground and ends up 1.30 m above the ground ? Change in gravitational potential energy:
Physics
1 answer:
den301095 [7]3 years ago
5 0

Answer: 324.135 kJ

Explanation:

Given

Mass of dropping is m=2250\ kg

The initial height of dropping is h_1=16\ m

The final height of dropping h_2=1.3\ m

Gravitational potential energy is the function of height i.e.

\Rightarrow \text{G.E.}=mgh

Change in Gravitational Energy is

\Rightarrow \Delta \text{G.E.}=mg(h_1-h_2)=2250\times 9.8\times (16-1.3)\\\\\Rightarrow \Delta \text{G.E.}=3,24,135\ J\approx 324.135\ kJ

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dmitriy555 [2]

Answer:

  • To not exceed the 5,000 characters, see each answer with its explanation below.

Explanation:

<u><em>Question 1.</em></u>

<em>A feather and a rock dropped at the same time from the same height would land at the same time when dropped by an astronaut on the moon</em>.

The shape and mass of the <em>feather</em> that falls vertically down, through the air, cause the resistance of the air, that opposes the gravitational attraction of the Earth, to be considerably large, compared to the case of the <em>rock</em>.

This not negligible resistance of the air causes a drag on the <em>feather</em> that makes it fall more slowly than the <em>rock</em>.

Only on the moon, the almost absence of atmosphere (air) would cause the <em>feather</em> not to be dragged and, therefore, its acceleration will be the same as that of the rock, and both will fall with the same speed and land at the same time (there is a video in the internet that shows this experiment of the astronaut on the moon).

<em><u>Question 2.</u></em>

<em>When the soccer ball is kicked, the action and reaction forces do not cancel each other out because the forces act on two different objects.</em>

The third law of Newton, also known as the law of action and reaction, states that whenever an object experience a force (action force) from a second object, the first object will exert a force of equal size and opposite in direction (reaction force) over the second object.

Hence, the action and reaction forces act over distinct objects. The ball feels the action force (the kick) and the foot feels the reaction force from the ball; the forces do not cancel each other because they act on two different objects.

<u><em>Question 3 </em></u>

<em>An object is in projectile motion if it is thrown with a horizontal push.</em>

The <em>projectile motion</em> is the combination of two motions: vertical motion and horizonal motion.

The vertical motion is ruled by the acceleration of gravity and will depend on both the inital vertical velocity and the acceleration of gravity.

The horizontal motion is ruled by the horizontal velocity. According with the inertia law (first law of Newton), in the absence of a horizontal force, only if the object departs with a horizontal velocity it will move horizontally.

Hence, an object can only be in projectile motion if it is thrown with a horizontal push.

<u><em>Question 4 </em></u>

Acceleration of an object

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  • decreases as the mass of the object increases.
  • is in the same direction as the force on the object.
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According to second law of Newton

          \vec{F}=m\times \vec{a}

This is force equals mass time acceleration.

The force is a vector and the acceleration is a vector, then since mass is a positive scalar magnitude, both force and acceleration have the same direction.

Also, it is clear from the definition of force that it is proportional to both, the mass and the acceleration, then if either mass or acceleration increases (while the other remains equal) the force will increase, and if either mass or acceleration decreases (while the other remains constant) the force will decrease.

Hence, all the options are correct.

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<em>The bowling ball will have more momentum because it has more mass.</em>

Momentum, P, is the product of the mass (m) and the velocity (v).

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It is clear that the mass of the bowling ball is greater than the mass of the golf ball.

Hence, since they are moving at the same velocity, the product of the mass of the bowling ball times the velocity is greater than the product of the golf mass times the same velocity.

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<em>An  </em><em><u>   orbit         </u></em><em> is formed when an object is falling around another object, rather than into it.</em>

An example of an object falling around another object is a rock falling in the Earth. The gravitaional pull of the Earth causes that the rock accelerates and falls vertically downward.

An example of an object falling around another object is a satellite around the Earth.

The Earth is continually pulling the satellite with vertical force and the inertia (initial velocity) of the satellite is causing it to move horizontally.

The combination of the gravitational force and the inertia cause the satellite to follow a projectile motion. When the satellite is to the correct distance it will not reach into the Earth but will continually move around the Earth following a circular path named orbit.

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<em>When the force of gravity is matched by the force of air resistance, an object can reach  </em><em><u>    terminal velocity      </u></em><em>(two word answer).</em>

When the force of gravity matches the force of air resistance, the net force on the falling object is zero.

Hence, as per the second Law of Newton, force equals the product of the mass and the acceleration, the acceleration is also zero.

Zero acceleration means constant velocity.

That constant velocity is a limit or maximum for the velocity that an object can reach when falling and is called terminal velocity.

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larisa [96]

Answer:

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Explanation:

This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.

Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point

                q_A = Q / 2

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Now sphere A touches sphere C, ending with half the charge

                q_A = ½ (Q / 2) = ¼ Q

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Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge

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