In 1920, after returning from Army service, he produced a successful model and in 1923 turned it over to the Northeast Electric Company of Rochester for development.
Answer:
80 ft/s
Explanation:
Use III equation of motion
V^2 = U^2 + 2g h
Here, U = 0, g = 32 ft/s^2, h = 100 ft
V^2 = 0 + 2 × 32 ×100
V^2 = 6400
V = 80 ft/s
Answer: 3 m.
Explanation:
Neglecting the mass of the seesaw, in order the seesaw to be balanced, the sum of the torques created by gravity acting on both children must be 0.
As we are asked to locate Jack at some distance from the fulcrum, we can take torques regarding the fulcrum, which is located at just in the middle of the length of the seesaw.
If we choose the counterclockwise direction as positive, we can write the torque equation as follows (assuming that Jill sits at the left end of the seesaw):
mJill* 5m -mJack* d = 0
60 kg*5 m -100 kg* d =0
Solving for d:
d = 3 m.
