The length decreased during the observation period.
2 answers:
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Answer:
Explanation:
Answer:
0.632 m
Explanation:
let the equilibrium separation between the charges is d and the angle made by string with the vertical is θ.
According to the diagram,
d = L Sinθ + L Sinθ = 2 L Sinθ .....(1)
Let T be the tension in the string.
Resolve the components of T.
T Sinθ = k q1 q2 / d^2
T Sinθ = k q1 q2 / (2LSinθ)² .....(2)
T Cosθ = mg .....(3)
Dividing equation (2) by equation (3), we get
tanθ = k q1 q2 / (4 L² Sin²θ x mg)
tan θ Sin²θ = k q1 q2 / (4 L² m g)
For small value of θ, tan θ = Sin θ
So,
Sin³θ = k q1 q2 / (4 L² m g)
Sin³θ = (9 x 10^9 x 0.785 x 10^-6 x 1.47 x 10^-6) / (4 x 0.5 x 0.5 x 4.20 x 10^-3 x 9.8)
Sin³θ = 0.2523
Sinθ = 0.632
θ = 39.2 degree
So, the separation between the two charges, d = 2 x L x Sin θ
d = 2 x 0.5 x 0.632 = 0.632 m
Answer:
Explanation:
a) using the energy conservation equation
mgh = 0.5mv^2 + 0.5Iω^2
I(moment of inertia) (basket ball) = (2/3)mr^2
mgh = 0.5mv^2 + 0.5( 2/3mr^2) ( v^2/r^2)
gh = 1/2v^2 + 1/3v^2
gh = v^2( 5/6)
v =
putting the values we get
solving for h( height)
h = 3.704 m apprx
b) velocity of solid cylinder
mgh = 0.5mv^2 + 0.5( mr^2/2)( v^2/r^2) where ( I ofcylinder = mr^2/2)
g*h = 1/2v^2 + 1/4v^2
g*h = 3/4v^2
putting the value of h and g we get
v= = 6.957 m/s apprx