Answer:
(a) Friction force = 50 N
(b) Work done by friction = 300 j
(c) Net work done = 0 j
Explanation:
We have given that the box is pulled by 6 meter so d = 6 m
Force applied on the box F = 60 N
We have have given that velocity is constant so acceleration will be zero
So to applied force will be utilized in balancing the friction force
So friction force 
Work done by friction force 
Work done by applied force 
So net work done = 300-300 = 0 j
Answer: both hoops have the same kinetic energy at the bottom of the incline.
Explanation:
If we assume no work done by non conservative forces (like friction) , the total mechanical energy must be conserved.
K1 + U1 = K2 + U2
If both hoops start from rest, and we choose the bottom of the incline to be the the zero reference level for gravitational potential energy, then
K1 = 0 and U2 = 0
⇒ ΔK = ΔU = m g. h
If both inclines have the same height, and both hoops have the same mass m, the change in kinetic energy, must be the same for both hoops.
Answer:
0.09 N
Explanation:
We are given that
Radius of disk,r=6 cm=
1 m=100 cm
B=1 T
Current,I=3 A
We have to find the frictional force at the rim between the stationary electrical contact and the rotating rim.





Torque due to friction

Where friction force=F


Substitute the values


Answer:
Wavelength is 0.5
Explanation:
To work it out, you divide Wave speed by the Frequency (24÷48=0.5)