Answer:
im not 100% sure i should be doing this but i found a website that will help a ton.
https://geo.libretexts.org/Courses/Gettysburg_College/Book%3A_An_Introduction_to_Geology_(Johnson_Affolter_Inkenbrandt_and_Mosher)/02%3A_Plate_Tectonics/2.01%3A_Alfred_Wegener%E2%80%99s_Continental_Drift_Hypothesis
Answer:
Part a)

Part b)

Explanation:
Part a)
In order to have same range for same initial speed we can say


so after comparing above we will have

so we have


Part b)
Time of flight for the first ball is given as



Now for other angle of projection time is given as


So here the time lag between two is given as



First, figure out what the angular speed of the merry-go-around is.
v ϖ = Using the formula for linear speed Since the angular speed is constant, there is no angular acceleration. Tangential acceleration is t a Radial acceleration is r a Thus the total acceleration is a = rω =1.2m×1.6rad / s =1.9m/ s = rα =1.2m×0rad / s2 = 0m/ s2 =rϖ2 =1.2m×(1.6rad / s)2 =3.1m/ s2 2 2 t r = a +a = 0+(3.1)2 =3.1m/ s2 1 4.0 rev
Answer:
A force of 75 N placed at 0.7 m on the meter stick.
Explanation:
The weight of the box is equal to:

The net torque is equal to zero and is equal to:

For a force with a value of 75 N that is placed at 0.70 m on the meter stick, it would produce a torque of 15 N m
If you replace that values in the equation:
