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Sati [7]
4 years ago
5

Two charged bodies exert a force 0.8N on each other in a medium of relative permittivity 0.45.what will be the new force when th

ey are placed in a medium of relative permittivity 0.9,assuming all other parameters remain unchanged.​

Physics
1 answer:
Damm [24]4 years ago
7 0

The electric force between charged bodies decreases by a factor equal to the relative permittivity, ie:

F ∝ 1/ε

F = electric force, ε = relative permittivity

Find the scaling factor of F due to the change in ε:

k = 1/(0.9/0.45) = 1/2 = 0.5

Multiply the original F by k to find the new F:

F = 0.8N×k

F = 0.8N×0.5

F = 0.4N

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d) 289.31 m

Explanation:

Energy provided by potential energy = mgh = m x 9.8x 200 sin10.5 = 357.18m

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Energy used by friction on plain surface = μmg x d.( dis distance covered on plain ) =.075x m x 9.8 xd = .735 m d

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4 0
3 years ago
A 0.45kg baseball is pitched towards home plate at 20 m/s. The ball is hit back towards the pitcher with a speed of 30 m/s. What
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4.5kgm/s

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Change in momentum is expressed as

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M is the mass

V is the final velocity

u is the initial velocity

Given

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v = 30m/s

u = 20m/s

Substitute

Change in momentum = 0.45(30-20)

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3 0
3 years ago
A bullet is fired horizontally from a handgun at a target 100.0 m away. If the initial speed of the bullet as it leaves the gun
Lera25 [3.4K]

Answer:

The distance is 0.53 m.

Explanation:

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We need to calculate the total time

Using formula of time

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Put the value into the formula

t=\dfrac{100.0}{300}

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s=ut+\dfrac{1}{2}gt^2

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