Work done = Potential Energy = (mg)h
mg = weight of the ball
= 1000 N * 2.5 m = 2500 Joules.
The work done during the 5 second is 2500 Joules.
Answer:
Electric field is a function 1/r^2
Explanation:
Answer:
μ = 0.125
Explanation:
To solve this problem, which is generally asked for the coefficient of friction, we will use the conservation of energy.
Let's start working on the ramp
starting point. Highest point of the ramp
Em₀ = U = m h y
final point. Lower part of the ramp, before entering the rough surface
= K = ½ m v²
as they indicate that there is no friction on the ramp
Em₀ = Em_{f}
m g y = ½ m v²
v = 
we calculate
v = √(2 9.8 0.25)
v = 2.21 m / s
in the rough part we use the relationship between work and kinetic energy
W = ΔK = K_{f} -K₀
as it stops the final kinetic energy is zero
W = -K₀
The work is done by the friction force, which opposes the movement
W = - fr x
friction force has the expression
fr = μ N
let's write Newton's second law for the vertical axis
N-W = 0
N = W = m g
we substitute
-μ m g x = - ½ m v²
μ = 
Let's calculate
μ = 
μ = 0.125
Answer:
5.78amps
Explanation:
Given data
Time t= 57 seconds
Charge Q= 330C
Current I= ??
The expression for the electric current is given as
Q= It
Substituting we have
330= I*57
I= 330/57
I=5.78 amps
Hence the current is 5.78amps
6. D
7. D
8. B
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