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2 years ago

A conducting sphere of radius 5. 0 cm carries a net charge of 7. 5 µc. what is the surface charge density on the sphere?

Physics

1 answer:

elena-s [515]2 years ago

3 0

The surface charge density on the sphere is 0.014C/ $m^{2}$ .

Given,

r=5.00cm=0.05m, q=7. 5 µc

We know surface charge density σ =q/V= $\frac{7.5*10^{-6} }{\frac{4}{3}\pi 0.05^{3} }$ =0.014C/ $m^{2}$ .

<h3>Surface charge density </h3>

It is always important to know how charge is moving in an electric field. These fields will also build up electric charges. Therefore, estimation of the surface charge density is essential for a number of applications. It is also necessary to calculate the surface charge density of an electric object using its volume and surface area. The amount of electric charge that has collected in a given field is measured by the surface charge density. Using the dimensions provided, it determines the amount of electric charge. Dimensions of the electric body could be in the form of length, area, or volume. In one, two, or three dimensions, according to electromagnetism, surface charge density is a measurement of the amount of electric charge present in a given volume of space.

A conducting sphere of radius 5. 0 cm carries a net charge of 7. 5 µc. what is the surface surface charge density on the sphere?

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Answer:

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Explanation:

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3 years ago

a 1-kg discus is thrown with a velocity of 19 m/s at an angle of 35 degrees from the vertical direction. calculate the vertical

nlexa [21]

Answer:

Vx = 10.9 m/s , Vy = 15.6 m/s

Explanation:

Given velocity V= 19 m/s

the angle 35 ° is taken from Y-axis so the angle with x-axis will be 90°-35° = 55°

θ = 55°

to Find Vx = ? and Vy= ?

Vx = V cos θ

Vx = 19 m/s × cos 55°

Vx = 10.9 m/s

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3 years ago

Read 2 more answers

A 2.4-kg ball falling vertically hits the floor with a speed of 2.5 m/s and rebounds with a speed of 1.5 m/s. What is the magnit

gayaneshka [121]

Answer:

9.6 Ns

Explanation:

Note: From newton's second law of motion,

Impulse = change in momentum

I = m(v-u).................. Equation 1

Where I = impulse, m = mass of the ball, v = final velocity, u = initial velocity.

Given: m = 2.4 kg, v = 2.5 m/s, u = -1.5 m/s (rebounds)

Substitute into equation 1

I = 2.4[2.5-(-1.5)]

I = 2.4(2.5+1.5)

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I = 9.6 Ns

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3 years ago

Susan, driving north at 53 mphmph , and Shawn, driving east at 63 mphmph , are approaching an intersection. Part A What is Shawn

mafiozo [28]

Answer:

Shawn's speed relative to Susan's speed = 10 mph

Resultant velocity = 82.32 mph

Explanation:

The given data :-

i) Susan driving in north and speed of Susan is ( v₁ ) = 53 mph.

ii) Shawn driving in east and speed of Shawn is ( v₂ ) = 63 mph.

iii) The speed of both Susan and Shawn is relative to earth.

iv) The angle between Susan in north and Shawn in east is 90°.

We have to find Shawn's speed relative to Susan's speed.

v₂₁ = v₂ - v₁ = 63 - 53 = 10 mph

Resultant velocity,

$v = \sqrt{v_{2} ^{2}+ v_{1} ^{2} } =\sqrt{63^{2} +53^{2} }$

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3 years ago

Astronauts use a centrifuge to simulate the acceleration of a rocket launch. The centrifuge takes 40.0 s to speed up from rest t

Vinvika [58]

Answer

Time period T = 1.50 s

time t = 40 s

r = 6.2 m

Angular speed ω = 2π/T

= $\dfrac{2\pi }{1.5}$

= 4.189 rad/s

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= $\dfrac{4.189}{40}$

= 0.105 rad/s²

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v = 2πr/T

= $\dfrac{2\pi \times 6.2}{1.5}$

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So centripetal acceleration.

a = $\dfrac{v^2}{r}$

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= 11.1 g

in combination with the gravitation acceleration.

$a_{total} = \sqrt{(11.1g)^2+g^2}$

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3 years ago