What is the magnification of an object that is 4.15 m in front of a camera that has an image position of 5.0 cm?
1 answer:
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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:
Efriction = 768.23 [kJ]
Explanation:
In order to solve this problem we must use the principle of energy conservation. Where it tells us that the energy of a system plus the work applied or performed by that system, will be equal to the energy in the final state. We have two states the initial at the time of the balloon jump and the final state when the parachutist lands.
We must identify the types of energy in each state, in the initial state there is only potential energy, since the reference level is in the ground, at the reference point the potential energy is zero. At the time of landing the parachutist will only have potential energy, since it reaches the reference level.
The friction force acts in the opposite direction to the movement, therefore it will have a negative sign.
where:
m = mass = 56 [kg]
h = elevation = 1400 [m]
v = velocity = 5.6 [m/s]