In order to develop this problem it is necessary to use the concepts related to the conservation of both potential cinematic as gravitational energy,


Where,
M = Mass of Earth
m = Mass of Object
v = Velocity
r = Radius
G = Gravitational universal constant
Our values are given as,



Replacing we have,




Therefore the speed of the object when striking the surface of earth is 4456 m/s
Answer:
471392.4 N
Explanation:
From the question,
Just before contact with the beam,
mgh = Fd.................... Equation 1
Where m = mass of the beam, g = acceleration due to gravity, h = height. F = average Force on the beam, d = distance.
make f the subject of the equation
F = mgh/d................ Equation 2
Given: m = 1900 kg, h = 4 m, d = 15.8 = 0.158 m
Constant: g = 9.8 m/s²
Substitute into equation 2
F = 1900(4)(9.8)/0.158
F = 471392.4 N
scientific notation is given as

here we know that
1 < a < 10 and
k = whole number
now the number will be
N = 299,790,000
here we know that

so we have
a = 2.99
k = 8
Answer:
D) multiply the force in the direction of motion by the distance the object moved
Explanation:
To calculate work done on an object, multiply the force in the direction of motion by the distance the object moved.