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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.)
Physics
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:

The coefficient of friction present between the roadway and the wheels of the truck is <u>0.833</u>.

Explanation:

Given:

Radius of the curve (R) = 150 m

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Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

F_c=\frac{mv^2}{R}

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

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N=mg

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Plug in the given values and solve for 'μ'. This gives,

\mu=\frac{(35\ m/s)^2}{(150\ m)(9.8\ m/s^2)}\\\\\mu=\frac{1225\ m^2/s^2}{1470\ m^2/s^2}\\\\\mu=0.833

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A heavy meterstick has a mass of 1 kg. When the meterstick is thrown like a spear past you, you measure its momentum to be 2mv.
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Answer:

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