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
1 answer:
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 = 12262 V/m
Hence, the electric field acting in the particle is equal to E = 12262 V/m
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Answer:
x = 76.5 m
Explanation:
Let's use Newton's second law at the point of contact between the wheel and the floor.
fr = m a
fr = miy N
N-W = 0
N = W
μ mg = m a
a = miu g
a = 0.600 9.8
a = 5.88 m / s²
Having the acceleration we can use the kinematic relationships to find the distance
² = v₀² + 2 a x
= 0
x = -v₀² / 2 a
Acceleration opposes the movement by which negative
x = - 30²/2 (-5.88)
x = 76.5 m