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Varvara68 [4.7K]

3 years ago

8

A proton starts from rest near the surface of a sheet of charge. It experiences a force of 14.0 microNewtons towards the sheet.

What is the surface charge density of the sheet? (Ignore the gravitational force on the proton.)

1 answer:

Naddika [18.5K]3 years ago

6 0

Answer:

The surface charge density of the sheet is 1547.516 C/m²

Explanation:

Given;

force experienced by the proton, F = 14.0 μN

Electric field near an infinite sheet with surface charge density (σ) is given as;

E = σ/2ε₀

The force of attraction the proton experienced towards the sheet;

F = Eq

F = q(σ/2ε₀) $= \frac{q \sigma}{2 \epsilon_o}$

where;

q is the charge of the proton

σ is the surface charge density of the sheet

ε₀ is the permittivity of free space

$F = \frac{q \sigma}{2 \epsilon_o} \\\\\sigma = \frac{2F \epsilon_o}{q} \\\\\sigma = \frac{2*14*10^{-6}*8.854*10^{-12}}{1.602*10^{-19}} = 1547.516 \ C/m^2$

Therefore, the surface charge density of the sheet is 1547.516 C/m²

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Answer:

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Explanation:

From the information given:

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$\Delta _{\omega} = \omega - 0 \\ \\ = 50 \ rad/s$

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$I= mr^2 \\ \\ I = (3 \times 0.2^2) \\ \\ I = 0.12 \ kgm^2$

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3 years ago

Which mixtures could be separated using evaporation? Check all that apply. fat in milk salt in ocean water mud in pond water but

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Answer:

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$V_w$ = Velocity of wind

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Subtracting the two equations we get

$V_a-V_w-V_a-V_w=460-540\\\Rightarrow -2V_w=-80\\\Rightarrow V_w=40\ km/h$

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$V_a-40=460\\\Rightarrow V_a =500\ km/h$

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3 years ago

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3 years ago

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jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

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Suppose we represent $(v_{stone})$ to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

$m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}$

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3 years ago