I NEED THIS A SOON AS POSSIBLE
2 answers:
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Answer:
10628.87 J
Explanation:
We are given that
Force applied =F=5592 N
Displacement=D=3.79 m
We have to find the work done in sliding the piano up the plank at a slow constant rate.
Work done=
The perpendicular component of force=
Work done =
Hence, the work done in sliding the piano up the plank at a slow constant rate=10628.87 J
Answer:
Part a: The electric potential of each sphere is 1.35x10⁸V
Part b: The electric field at the surface of sphere 1 and 2 is 2.25x10⁹ N/C and 6.75x10⁹ N/C respectively
Explanation:
As the complete question is not given, the similar question is attached herewith. The values are used as indicated in the given question
Let r_1 = 6 cm=0.06 m
r2 = 2 cm = 0.02 m
Q = 1.2 mC
Let q1 and q2 are the charges on each sphere.
q1 + q2 = 1.2 mC -------(1)
In the equilibrium, V1 = V2
k*q1/r1 = k*q2/r2
q1/0.06 = q2/0.02
q1/q2 = 0.06/0.02
q1/q2 = 3 ---------(2)
On solving equation 1 and 2
we get
q1 = 0.9 mC
q2 = 0.3 mC
So
V1 = k*q1/r1 = (9*10^9*0.9*10^-3)/0.06 = 1.35*10^8 Volts
V2 = k*q2/r2 = 9*10^9*0.3*10^-3/0.02 = 1.35*10^8 Volts
So the electric potential of each sphere is 1.35x10⁸V
Part b
Now the electric potential is given as
E1 = k*q1/r1^2 = 9*10^9*0.9*10^-3/0.06^2 = 2.25*10^9 N/C
E2 = k*q2/r2 = 9*10^9*0.3*10^-3/0.02^2 = 6.75*10^9 N/C
So the electric field at the surface of sphere 1 and 2 is 2.25x10⁹ N/C and 6.75x10⁹ N/C respectively
