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Alenkasestr [34]
2 years ago
5

A 4.0 µC point charge and a 3.0 µC point charge are a distance L apart. Where should a third point charge be placed so that the

electric force on that third charge is zero? (Use the following as necessary: L.)
Physics
1 answer:
Rudik [331]2 years ago
3 0

Answer:

Q must be placed at 0.53 L

Explanation:

Given  data:

q_1 = 4.0 μC , q_2 = 3.0μC

Distance between charge is L

third charge q be placed at  distance x cm from q1

The force by charge q_1 due to q is

F1 = \frac{k q q_1}{x^2}

F1 = \frac{k q ( 4.0 μC )}{ x^2}                  ----1

The force by charge q_2 due to q is

F2 =  \frac{k q q_2}{(L-x)^2}

F2 = \frac{kq (3.0 μC)}{(L-x)^2}                   --2

we know that net electric force is equal to zero

F_1 = F_2

\frac{k q ( 4.0 μC )}{x^2}   =\frac{k q ( 3.0 μC )}{(l-x)^2}

\frac{4}{3}*(L-x)^2 = x^2

x = \sqrt{\frac{4}{3}*(L - x)

L-x = \frac{x}{1.15}

L = x + \frac{x}{1.15} = 1.86 x

x = 0.53 L

Q must be placed at 0.53 L

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Question 2
Ksivusya [100]

5.77 ×10^1^4 Hz is the green photon's frequency .

The distance between similar points (adjacent crests) in adjacent cycles of a waveform signal that is propagated in space is known as the wavelength. A wave's wavelength is often measured in meters (m), centimeters (cm), or millimeters (mm) (mm). The relationship between frequency and wavelength is inverse.

<h3>Given:</h3>

Wavelength of green light = 520 nm

f = c / λ

where, f = Frequency

            c = Speed of light = 3 × 10^8 m/s

            λ = Wavelength of light

∴ f = c / λ

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    = 5.77 ×10^1^4 Hz

Therefore,  5.77 ×10^1^4 Hz is the green photon's frequency .

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4 0
1 year ago
A 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees. If the car is traveling at 95 km/h, will a frictio
Mariulka [41]

Answer:

Yes. Towards the center. 8210 N.

Explanation:

Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.

In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.

The net force is equal to F = \frac{mv^2}{R} = \frac{1100\times (26.3)^2}{68} = 1.1\times 10^4~N

Note that 95 km/h is equal to 26.3 m/s.

This is the centripetal force and equal to the x-component of the applied force.

F = mg\sin(16) = 1100(9.8)\sin(16) = 2.97\times10^3

As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.

The amount of the friction force should be 8.21\times 10^3~N

Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.

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3 years ago
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A believe that’s called a reference point.
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