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Natalija [7]
3 years ago
10

A particle with a charge of +4.0 μC has a mass of 5.0 g. What magnitude electric field directed upward will exactly balance the

weight of the particle?
Physics
1 answer:
aleksley [76]3 years ago
3 0

Answer:

E = 12262 V/m

Explanation:

given,

Charge of the particle, q = +4.0 μC

mass of the ball, m = 5 g

Electric field, E = ?

Force due to weight

F = m g

Force due to electric field

F =q E

To balance the weight of particle both the forces must be equal

Electric field exerted on the ball will be equal to

m g = q E

E = \dfrac{m\ g}{q}

E = \dfrac{5\times 10^{-3}\times 9.81}{4\times 10^{-6}}

E = 12262 V/m

Hence, the electric field acting in the particle is equal to E = 12262 V/m

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The acceleration of the particle as a function of time t is
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To solve the problem, now we just have to substitute t=5 into x(t) and v(t) to find the position and the velocity of the particle at t=5.

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A person is sitting with one leg outstretched and stationary, so that it makes an angle of θ = 27.5° with the horizontal, as the
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here we will use the torque balance about the knee joint

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\tau_g = \tau_m

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