Am = 94.2 kg object is released from rest at a distance h = 1.15134 r above the earth’s surface. the acceleration of gravity is 9.8 m/s 2 . for the earth, re = 6.38 × 106 m, m = 5.98 × 1024 kg. the gravitational acceleration at the surface of the earth is g = 9.8 m/s 2 . find the speed of the object when it strikes the earth’s surface. neglect any atmospheric friction. caution: you must take into account that the gravitational acceleration depends on distance between the object and the center of the earth. answer in units of m/s.

v= 4055.08m/s

Explanation:

This is a problem that must be addressed through the laws of classical mechanics that concern Potential Gravitational Energy.

We know for definition that,

We must find the highest point and the lowest point to identify the change in energy, so

Point a)

The problem tells us that an object is dropped at a distance of h = 1.15134R over the earth.

That is to say that the energy of that object is equal to,

Point B )

We now use the average radius distance from the earth.

Then,

By the law of conservation of energy we know that,

clearing v,

Therefore the speed of the object when it strikes the Earth’s surface is 4055.08m/s

adding a(n) to an electric circuit would increase the magnetic field around a wire;

b. solenoid

a solenoid is a type of electromagnet whose purpose is to generate a controlled magnetic field.

if the purpose of the solenoid is instead to impede changes in the electric current, a solenoid can be more specifically classified as an inductor rather than an electromagnet.

answer; equivalent to a one(1) horsepower motor;