Answer:
6495.19 Joule
Explanation:
F = Weight of the crate = 250 N
d = Distance the cart is pushed = 30 m
θ = Angle of inclination = 60°
The weight of the crate will be resloved into two components
Fdsinθ and Fdcosθ
Work done by the force of gravity is
W = Fdsinθ
⇒W = 250×30×sin60
⇒W = 6495.19 Joule
∴ The work done by the force of gravity is 6495.19 Joule
Answer:
R = 710.7N
L = 67.689 N
During gravity fall L = R = 0 N
Explanation:
So the acceleration that the elevator is acting on the woman (and the package) in order to result in a net acceleration of 0.15g is
g + 0.15g = 1.15g
The force R that the elevator exerts on her feet would be product of acceleration and total mass (Newton's 2nd law):
a(m + M) = 1.15g(57 + 6) = 1.15*9.81*63 = 710.7N
The force L that she exerts on the package would be:
am = 1.15g *6 = 1.15*9.81*6 = 67.689N
When the system is falling, all have a net acceleration of g. So the acceleration that the elevator exerts on the woman (and the package) is 0, and so are the forces L and R.
This is true!!
Good luck hope this helped!
Answer with Explanation:
We are given that
Mass of box=35 kg
Coefficient of static friction between box and truck bed=0.202
Acceleration due to gravity=
a.We have to find the force by which the box accelerates forward.
Force by which box accelerates=
Force by which box accelerates=
b.We have to find the maximum acceleration can the truck have before the box slides.
Force =friction force
Hence, the truck can have maximum acceleration before the box slide=
Wouldn’t it be 1. resistance is high compared to the voltage and with less resistance, higher current
