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Aleonysh [2.5K]
1 year ago
15

If a point charge is located at the center of a cylinder and the electric flux leaving one end of the cylinder is 20% of the tot

al flux leaving the cylinder, what portion of the flux leaves the curved surface of the cylinder
Physics
1 answer:
zheka24 [161]1 year ago
5 0

The portion of the flux leaves the curved surface of the cylinder is 60%.

<h3 /><h3>What are electrons?</h3>

The electrons are the spinning objects around the nucleus of the atom of the element in an orbit.

If a point charge is located at the center of a cylinder and the electric flux leaving one end of the cylinder is 20% of the total flux leaving the cylinder.

If 20% of the flux leave from one end, then another 20% will leave from another end.

So, the net flux through curved surface is

100 -20 -20 = 60%

Thus, the total flux  leaves the curved surface of the cylinder is 60%

Learn more about electrons.

brainly.com/question/1255220

#SPJ1

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In a double‑slit interference experiment, the wavelength is lambda=487 nm , the slit separation is d=0.200 mm , and the screen i
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Answer:

Δx = 4.68 x 10⁻³ m = 4.68 mm

Explanation:

The distance between the consecutive maxima, in Young's Double Slit Experiment is given bu the following formula:

Δx = λD/d

So, the distance between the eighth order maximum and the fourth order maximum on the screen will be given as:

Δx = 4λD/d

where,

Δx = distance between eighth order maximum and fourth order maximum=?

λ = wavelength = 487 nm = 4.87 x 10⁻⁷ m

d = slit separation = 0.2 mm = 2 x 10⁻⁴ m

D = Distance between slits and screen = 48 cm = 0.48 m

Therefore,

Δx = (4)(4.87 x 10⁻⁷ m)(0.48 m)/(2 x 10⁻⁴ m)

<u>Δx = 4.68 x 10⁻³ m = 4.68 mm</u>

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What statement best describes his motion as he jogs around the curved part of the track?
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A powerful searchlight shines on a man. The man's cross-sectional area is 0.500m2 perpendicular to the light beam, and the inten
babymother [125]

Answer:

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N

(b) the force the light beam exerts is much too small to be felt by the man.

Explanation:

Given;

cross-sectional area of the man, A = 0.500m²

intensity of light, I = 35.5kW/m²

If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;

P = I/c

where;

P is the pressure of the incident light

I is the intensity of the incident light

c is the speed of light

P = \frac{I}{c} =\frac{35500}{3*10^8} = 1.18*10^{-4} \ N/m^2

F = PA

where;

F is the force of the incident light on the man

P is the pressure of the incident light on the man

A is the cross-sectional area of the man

F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N

Therefore, the force the light beam exerts is much too small to be felt by the man.

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