F = M A
Force = (mass) x (acceleration)
= (1,650 kg) (4 m/s²) = 6,600 kg-m/s² = <em>6,600 Newtons</em>
Answer:
pressure= force/area
A solid resting on a horizontal surfaceexerts a normal contactforce equals to its weight. The pressure of the solid on the surface depends on the area of contact. (b) the area of contact between the two surfaces. The greater the force or the smaller the area the greater the pressure.
Answer:
Option C
Explanation:
Given that
Motor force is 250 N
Force of friction is 750 N
Weight is 8500 N
And, the normal force is 8500 N
Now based on the above information
Here length of the rector shows the relative magnitude forward force i.e. 250 N i..e lower than the frictional force i.e. backward and weight i.e. 8500 would be equivalent to the normal force
Answer:
vi = 4.77 ft/s
Explanation:
Given:
- The radius of the surface R = 1.45 ft
- The Angle at which the the sphere leaves
- Initial velocity vi
- Final velocity vf
Find:
Determine the sphere's initial speed.
Solution:
- Newton's second law of motion in centripetal direction is given as:
m*g*cos(θ) - N = m*v^2 / R
Where, m: mass of sphere
g: Gravitational Acceleration
θ: Angle with the vertical
N: Normal contact force.
- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:
m*g*cos(θ) - 0 = m*vf^2 / R
g*cos(θ) = vf^2 / R
vf^2 = R*g*cos(θ)
vf^2 = 1.45*32.2*cos(34)
vf^2 = 38.708 ft/s
- Using conservation of energy for initial release point and point where sphere leaves cylinder:
ΔK.E = ΔP.E
0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))
( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))
vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))
vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))
vi^2 = 22.744
vi = 4.77 ft/s
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
