Answer:
35.3 N
Explanation:
U = 0, V = 0.61 m/s, s = 0.39 m
Let a be the acceleration.
Use third equation of motion
V^2 = u^2 + 2 as
0.61 × 0.61 = 0 + 2 × a × 0.39
a = 0.477 m/s^2
Force = mass × acceleration
F = 74 × 0.477 = 35.3 N
Answer:
acceleration a = 1.04 m/s2
Explanation:
Assume the train has a speed of 23m/s when the last car passes the railway workers. Once this happens the last car would have traveled a total distance of the 180m distance between the railway worker standing 180 m from where the front of the train started plus the 75m distance from the first car to the last car:
s = 75 + 180 = 255 m
We can use the following equation of motion to find out the distance traveled by the car:
where v = 23 m/s is the velocity of the car when it passes the worker,
= 0m/s is the initial velocity of the car when it starts, a m/s2 is the acceleration, which we are looking for.



Your body continues to move unless stopped by the seatbelt. An object in motion will remain in motion. Since your body was already moving it will continue to.
Answer:
1.3823 rad/s
20.7345 m/s
28.66129935 m/s²

2006.29095 N radially outward
Explanation:
r = Radius = 15 m
m = Mass of person = 70 kg
g = Acceleration due to gravity = 9.81 m/s²
Angular velocity is given by

Angular velocity is 1.3823 rad/s
Linear velocity is given by

The linear velocity is 20.7345 m/s
Centripetal acceleration is given by

The centripetal acceleration is 28.66129935 m/s²
Acceleration in terms of g


Centripetal force is given by

The centripetal force is 2006.29095 N radially outward
The torque will be experienced when the centrifuge is speeding up of slowing down i.e., when it is accelerating and decelerating.
Answer:
Explanation:
1. FALL PROTECTION-GENERAL REQUIREMENTS (29 CFR 1926.501) 6,010 VIOLATIONS
2. HAZARD COMMUNICATION (29 CFR 1910.1200). 3,671
3. SCAFFOLDING (29 CFR 1926.451). 2,813