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Rina8888 [55]
2 years ago
9

An infinitely long cylinder of radius R has linear charge density λ. The potential on the surface of the cylinder is V0, and the

electric field outside the cylinder is Er=λ2πϵ0r. Part A Find the potential relative to the surface at a point that is distance r from the axis, assuming r>R. Express your answer in terms of the variables λ, r, R, V0, and appropriate constants. Vr = nothing

Physics
1 answer:
tamaranim1 [39]2 years ago
5 0

Answer:

V = V_0 - (lamda)/(2pi(epsilon_0))*ln(R/r)

Explanation:

Attached is the full solution

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Zamir and Talia raced through a maze. Zamir walked 2 m north, 2 m east, 4 m south, 2 m east, 4 m north, 2 m east, 3 m south, 4 m
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Answer : Zamir's displacement and Talia's displacement is equal.

Explanation :

Displacement is explained to be the changing position of an object.

Zamir covers total distance 27 m and Talia covers total distance 19 m but  Zamir's initial and final position and Talia's initial and final position is same.

So, we can say that Zamir's displacement and Talia's displacement is equal.

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On a guitar, the lowest toned string is usually strung to the E note, which produces sound at 82.4 82.4 Hz. The diameter of E gu
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The complete and comprehensive solution is attached.

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1 year ago
Onur drops a basketball from a height of 10\,\text{m}10m10, start text, m, end text on Mars, where the acceleration due to gravi
Doss [256]

Answer:

Explanation:

Given that,

Basket ball is drop from height

H=10m

It is dropped on planet mass

And the acceleration due to gravity on Mars is given as

g= 3.7m/s²

Time taken for the ball to reach the ground

Initial velocity of the body is zero

u=0m/s

Using equation of motion: free fall

H = ut + ½gt²

10 = 0•t + ½ × 3.7 ×t²

10 = 0 + 1.85t²

10 = 1.85t²

Then, t² =10/1.85

t² = 5.405

t = √ 5.405

t = 2.325seconds

So the time the ball spend on the air before reaching the ground is 2.325 seconds

5 0
2 years ago
If you are driving 72 km/h along a straight road and you look to the side for 4.0 s, how far do you travel during this inattenti
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We know that speed equals distance between time. Therefore to find the distance we have that d = V * t. Substituting the values d = (72 Km / h) * (1h / 3600s) * (4.0 s) = 0.08Km.Therefore during this inattentive period traveled a distance of 0.08Km
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A 50.-kilogram rock rolls off the edge of a cliff. if it is traveling at a speed of 24.2 m/s when it hits the ground, what is th
ElenaW [278]

The correct answer to the question is : 29.88 m.

EXPLANATION :

As per the question, the mass of the rock m = 50 Kg.

The rock is rolling off the edges of the cliff.

The final velocity of the rock when it hits the ground v = 24 .2 m/s.

Let the height of the cliff is h.

The potential energy gained by the rock at the top of the cliff = mgh.

Here, g is known as acceleration due to gravity, and g = 9.8\ m/s^2

When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.

When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.

The kinetic energy of the rock when it touches the ground is given as -

                Kinetic energy K.E = \frac{1}{2}mv^2.

From above we know that -

   Kinetic energy at the bottom of the cliff = potential energy at a height h

                 \frac{1}{2}mv^2=\ mgh

                ⇒ v^2=\ 2gh

                ⇒ h=\ \frac{v^2}{2g}

                ⇒ h=\ \frac{(24.2)^2}{2\times 9.8}

                ⇒ h=\ 29.88\ m

Hence, the height of the cliff is 29.88 m

             


5 0
2 years ago
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