Answer:
h' = 603.08 m
Explanation:
First, we will calculate the initial velocity of the pellet on the surface of Earth by using third equation of motion:
2gh = Vf² - Vi²
where,
g = acceleration due to gravity on the surface of earth = - 9.8 m/s² (negative sign due to upward motion)
h = height of pellet = 100 m
Vf = final velocity of pellet = 0 m/s (since, pellet will momentarily stop at highest point)
Vi = Initial Velocity of Pellet = ?
Therefore,
(2)(-9.8 m/s²)(100 m) = (0 m/s)² - Vi²
Vi = √(1960 m²/s²)
Vi = 44.27 m/s
Now, we use this equation at the surface of moon with same initial velocity:
2g'h' = Vf² - Vi²
where,
g' = acceleration due to gravity on the surface of moon = 1.625 m/s²
h' = maximum height gained by pellet on moon = ?
Therefore,
2(1.625 m/s²)h' = (44.27 m/s)² - (0 m/s)²
h' = (1960 m²/s²)/(3.25 m/s²)
<u>h' = 603.08 m</u>
Answer: I maybe wrong but i'm pretty sure its C) Kinetic energy
Answer:
B) Friction
Explanation:
Friction is a force that acts when an object is sliding along a surface. Microscopically, this force is due to the fact that the two surfaces are not perfectly smooth, but they have "imperfections" that cause a force that opposes the motion of the object.
For an object sliding on a flat surface, the force of friction has magnitude:

where
is the coefficient of kinetic friction
m is the mass of the object
g is the acceleration of gravity
The direction of the force of friction is always opposite to the direction of motion of the object.
In reality, friction also acts if the object is at rest and it is pushed by a force; in this case, we talk about static friction, and its magnitude is

where
is called coefficient of static friction, and it is generally larger than the coefficient of kinetic friction.
Answer:
Option C. 5,000 kg m/s
Explanation:
<u>Linear Momentum on a System of Particles
</u>
Is defined as the sum of the momenta of each particles in a determined moment. The individual momentum is the product of the mass of the particle by its speed
P=mv
The question refers to an 100 kg object traveling at 50 m/s who collides with another object of 50 kg object initially at rest. We compute the moments of each object


The sum of the momenta of both objects prior to the collision is

