When the car speeds up, slows down, or goes around a curve,
passengers need a force applied to them to make them do the
same thing, otherwise they won't keep up with the car.
The force on the passenger is applied by means of friction between
the upholstery and the seat of his pants, and also by the seat-back
or his seat-belt.
Answer:
160.75 N
Explanation:
The downward velocity has no effect on the force situation, it is only changes in velocity (plus, of course, gravity, which is always there) that require a force. At constant velocity, the bottom spring s_3 is supporting its mass m_3 to balance gravity.
As the elevator slows, though, it also ends up slowing down the spring arrangement, too. However, because the stretching takes time, it means that some damped harmonic motion will be set up in the spring chain.
When the motion has finally damped out, the net force the bottom spring s3 exerts on m3 has two components--that of gravity and of the deceleration of the elevator:
F_3net = m3 * (g + a) = 10.5×(9.81+5.5)= 10.5×15.31= 160.75 N
Answer:
Angular acceleration will be 
Explanation:
We have given that mass m = 0.18 kg
Radius r = 0.32 m
Initial angular velocity 
And final angular velocity 
Time is given as t = 8 sec
From equation of motion
We know that 


So angular acceleration will be 
Answer:
Explanation:
position of centre of mass of door from surface of water
= 10 + 1.1 / 2
= 10.55 m
Pressure on centre of mass
atmospheric pressure + pressure due to water column
10 ⁵ + hdg
= 10⁵ + 10.55 x 1000 x 9.8
= 2.0339 x 10⁵ Pa
the net force acting on the door (normal to its surface)
= pressure at the centre x area of the door
= .9 x 1.1 x 2.0339 x 10⁵
= 2.01356 x 10⁵ N
pressure centre will be at 10.55 m below the surface.
When the car is filled with air or it is filled with water , in both the cases pressure centre will lie at the centre of the car .
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