When an elevator is accelerating downward, the normal force is equal to mg-ma (hence you feel a little lighter when accelerating downwards)
Therefore, the upward force of the elevator floor on the person must be less than 750N
Part A:
For this part we’re assuming all the kinetic energy of the moving bumper car is converted into elastic potential energy in the spring since the car is brought to rest. Therefore you can find the total kinetic energy to get your answer:
KE = ½ mv^2
KE = ½ (200)(8)^2
KE = 6400 J
Part B:
Now you can use Hooke’s law to find the force:
F = kx
F = (5000)(0.2)
F = 1000 N
Answer:
Explanation:
The magnitude of the net force exerted on q is known, we have the values and positions for 
and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted by on q. Then we can know the magnitude of the force exerted by 
about q, finally this will allow us to know the magnitude of
exerts a force on q in +y direction, and exerts a force on q in -y direction.
The net force on q is:
Rewriting for :
It was formed by ancient volcanic eruptions.
As it is given that the air bag deploy in time
total distance moved by the front face of the bag
Now we will use kinematics to find the acceleration
now as we know that
so we have
so the acceleration is 400g for the front surface of balloon
