Answer:
2.32 s
Explanation:
Using the equation of motion,
s = ut+g't²/2............................ Equation 1
Where s = distance, u = initial velocity, g' = acceleration due to gravity of the moon, t = time.
Note: Since Onur drops the basket ball from a height, u = 0 m/s
Then,
s = g't²/2
make t the subject of the equation,
t = √(2s/g')...................... Equation 2
Given: s = 10 m, g' = 3.7 m/s²
Substitute this value into equation 2
t = √(2×10/3.7)
t = √(20/3.7)
t = √(5.405)
t = 2.32 s.
Answer:
Is the equation for Ec=1/2 m(Dv)^2 where Dv is the difference between the angular speed & the areolar speed?
Can't really plot a graph here for question 1.
2a) The car speeds up from A to B. The car travels at a constant speed from B to C. The car slows down to a stop from C to D.
b) From the graph, at 10 seconds, the car is moving at 20 m/s.
