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Yuki888 [10]
3 years ago
11

Common static electricity involves charges ranging from nanocoulombs to microcoulombs. How many electrons must be removed from a

neutral object in order to leave a net charge of 9.7 μC?
Physics
1 answer:
madam [21]3 years ago
3 0

Answer:

6.0625 x 10^13

Explanation:

Charge of one electron, e = 1.6 x 10^-19 C

Total charge, Q = 9.7 μ C = 9.7 x 10^-6 C

The number of electrons is given by

n = Total charge / charge of one electron

n = Q / e

n = (9.7 x 10^-6) / (1.6 x 10^-19)

n = 6.0625 x 10^13

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Answer:

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Formula

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Determine the magnitude of the resultant force acting on a 1.5 −kg particle at the instant t=2 s, if the particle is moving alon
Phoenix [80]

Answer:

F = 63N

Explanation:

M= 1.5kg , t= 2s, r = (2t + 10)m and

Θ = (1.5t² - 6t).

magnitude of the resultant force acting on 1.5kg = ?

Force acting on the mass =

∑Fr =MAr

Fr = m(∇r² - rθ²) ..........equation (i)

∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)

The horizontal path is defined as

r = (2t + 10)

dr/dt = 2, d²r/dt² = 0

Angle Θ is defined by

θ = (1.5t² - 6t)

dθ/dt = 3t, d²θ/dt² = 3

at t = 2

r = (2t + 10) = (2*(2) +10) = 14

but dr/dt = 2m/s and d²r/dt² = 0m/s

θ = (1.5(2)² - 6(2) ) = -6rads

dθ/dt =3(2) - 6 = 0rads

d²θ/dt = 3rad/s²

substituting equation i into equation ii,

Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)

∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]

∑F = 1.5(14*3+0) = 63N

F = √(Fr² +FΘ²) = √(0² + 63²) = 63N

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3 years ago
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