Given :
A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.
To Find :
The coefficient of static friction between the box and the plane.
Solution :
Vertical component of force :
Horizontal component of force(Normal reaction) :
Since, box is on the verge of slipping :
Therefore, the coefficient of static friction between the box and the plane is 1.07.
Hence, this is the required solution.
Answer:
Work= -7.68×10⁻¹⁴J
Explanation:
Given data
q₁=q₂=1.6×10⁻¹⁹C
r₁=2.00×10⁻¹⁰m
r₂=3.00×10⁻¹⁵m
To find
Work
Solution
The work done on the charge is equal to difference in potential energy
W=ΔU
Answer:
B. 2 meters.
Explanation:
To rotate the capstan a certain amount of torque is required, and if each sailor applies a force at a distance
from the center, then for two sailors the total torque will be
;
therefore, for one sailor to apply the same torque it must be that the torque he applies must be equal to the torque that the two sailors applied:
which gives
.
and since ,
which is choice B.