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Varvara68 [4.7K]
2 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]2 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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attashe74 [19]

The normal force acting on the object is 500 N in the upward direction

<u>Explanation:</u>

As George is applying a downward force, the normal force will be in the upward direction. The normal force will be exerted due to the acceleration due to gravity exerted on the object.

So, as per Newton's second law, the normal force acting on the object can be measured by the product of mass of the object and the acceleration due to gravity acting on the object.

But as the acceleration due to gravity is a downward acting acceleration and the normal force is a upward acting force, so the acceleration will be having a negative sign in the formula.

Normal\ force = Mass \times Acceleration\ due\ to\ gravity

Here, acceleration due to gravity g = -10 m/s² and mass is given as 50 kg, then

Normal force = 50 × (-10) = -500 N

So, the normal force acting on the object is 500 N in the upward direction.

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3 years ago
An airplane is moving at 350 km/hr. If a bomb is
Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final jeight

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb'e initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's fina velocity

Knowing this, let's begin with the answers:

<h3>b) Time</h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity</h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

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Answer:

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Explanation:

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