5.00kg
random = abcdefghijklmnopqrstuvwxyz
About 21c because it also depends on the weather outside
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:
Explanation:
The concept of elastic and inelastic demand is applied.
for an elastic demand, the elasticity must be greater than 1 and for an Inelastic demand, the elasticity must be less than 1.
The steps and appropriate calculation is as shown in the attached file.
The spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is

Which means that the frequency is

and the angular frequency is

In a spring-mass system, the maximum velocity of the object is given by

where A is the amplitude of the oscillation. In our problem, the amplitude of the motion corresponds to the initial displacement of the object (A=0.500 m), therefore the maximum velocity is