Answer:
zero
Explanation:
For a solid conducting sphere, charges are present on the surface of the sphere due to a phenomenon known as electrostatic sheilding. This affects the charge present in the body and makes it zero. However, the electrostatic potential appears to be equal to the whole present point that shows on the surface. The surface of a spherical conducting solid sphere is known as an equipotential surface. Thus, the potential difference between the two opposite points on the surface of the sphere will also be zero.
In this collision, 1/2 of the initial kinetic energy of the first glider is converted into thermal energy.
<h3>In plain English, what is kinetic energy?</h3>
An object's strength as a result ofstrength an object has as a result or motion is known as kinetic energy. Toorder to accelerate an object, a force must be applied. Applying force requires effort on our part. When the work is done, power is transported to the thing, which causes it to move at the athe new, constant pace.
<h3>What does kinetic energy mean, or what are some instances?</h3>
The motion energy is known as kinetic energy, and it is manifested when a particle, object, or group if particles moves. Any moving object uses kinetic energy, including people walking, baseballs being thrown, food falling from tables, and charged particles in electric fields.
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The average power produced by the rocket is the product of force exerted by the rocket and the velocity of the rocket.
<h3>Average power produced by the rocket</h3>
The average power produced by the rocket is calculated as follows;
P = FV
where;
- P is the average power
- F is the force exerted by the rocket
- V is the velocity of the rocket
Thus, the average power produced by the rocket is the product of force exerted by the rocket and the velocity of the rocket.
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Kinetic energy of the rock just before it hits the ground=KE=33000 J
Explanation:
Weight= 2200N
mg=2200
m(9.8)=2200
m=224.5 kg
initial velocity=0
final velocity =V
using kinematic equation V²=Vi²+2gh
V²=0+2 (9.8)(15)
V=17.1 m/s
now kinetic energy= 1/2 mV²
KE= 1/2 (224.5)(17.1)²
KE=33000 J
Thus the kinetic energy of the rock just before it hits the ground=33000 J
All of the observations except "powerful gravitational field" are consistent with the current theory of black holes.
The gavitational field of a black hole is thought to be no different than that of an ordinary star with the same mass.
