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Furkat [3]
3 years ago
12

A positive charge is moved from point A to point B along an equipotential surface. How much work is performed or required in mov

ing the charge? A positive charge is moved from point A to point B along an equipotential surface. How much work is performed or required in moving the charge? Work is required in moving the positive charge from point A to point B. No work is performed or required in moving the positive charge from point A to point B. Work is performed in moving the positive charge from point A to point B. Work is both performed and required in moving the charge from point A to point B. SubmitRequest Answer Provide Feedback Next
Physics
1 answer:
Nitella [24]3 years ago
5 0

Answer:

No work is performed or required in moving the positive charge from point A to point B.

Explanation:

Lets take

Q= Positive charge which move from  point A to point B along

Voltage difference,ΔV =V₁ - V₂  

The work done

W = Q . ΔV

Given that  charge is moved from point A to point B along an equipotential surface.It means that voltage  difference is zero.

ΔV = 0

So

W = Q . ΔV

W = Q x 0

W= 0 J

So work is zero.

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Nookie1986 [14]

Explanation:

First of all we calculate the volume which is 20m×10m×8m= 1600m^3 and then multiply the density by the volume to get the mass which will equal to 99200kg therefore the weight will be

<em><u>Weight</u></em>

mass × acceleration due to gravity

=99200kg × 10m^s^2

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7 0
3 years ago
The drag on a pitched baseball can be surprisingly large. Suppose a 145 g baseball with a diameter of 7.4 cm has an initial spee
kupik [55]

Answer:

<h2>Part A)</h2><h2>Acceleration of the ball is 10.1 m/s/s</h2><h2>Part B)</h2><h2>the final speed of the ball is given as</h2><h2>v_f = 35.3 m/s</h2>

Explanation:

Part a)

As we know that drag force is given as

F = \frac{C_d \rho A v^2}{2}

C_d = 0.35

A = \frac{\pi d^2}{4}

A = \frac{\pi(0.074)^2}{4}

A = 4.3 \times 10^{-3} m^2

v = 40.2 m/s

so we have

F = \frac{0.35\times 1.2 (4.3 \times 10^{-3})(40.2)^2}{2}

F = 1.46 N

So acceleration of the ball is

a = \frac{F}{m}

a = \frac{1.46}{0.145}

a = 10.1 m/s^2

Part B)

As per kinematics we know that

v_f^2 - v_i^2 = 2 a d

v_f^2 - 40.2^2 = 2(-10.1)(18.4)

v_f = 35.3 m/s

4 0
3 years ago
Tarik winds a small paper tube uniformly with 183 turns 183 turns of thin wire to form a solenoid. The tube's diameter is 9.49 m
Rufina [12.5K]
<h2>Answer:</h2>

143μH

<h2>Explanation:</h2>

The inductance (L) of a coil wire (e.g solenoid) is given by;

L = μ₀N²A / l                 --------------(i)

Where;

l = the length of the solenoid

A = cross-sectional area of the solenoid

N= number of turns of the solenoid

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

<em>From the question;</em>

N = 183 turns

l = 2.09cm = 0.0209m

diameter, d = 9.49mm = 0.00949m

<em>But;</em>

A = π d² / 4                     [Take π = 3.142 and substitute d = 0.00949m]

A = 3.142 x 0.00949² / 4

A = 7.1 x 10⁻⁵m²

<em>Substitute these values into equation (i) as follows;</em>

L = 4π x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209           [Take π = 3.142]

L = 4(3.142) x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209

L = 143 x 10⁻⁶ H

L = 143 μH

Therefore the inductance in microhenrys of the Tarik's solenoid is 143

6 0
3 years ago
What were some major reasons for the american television from analog broadcast to digital broadcast?
Vika [28.1K]

Answer: It frees up valuable portions of the broadcast spectrum, it has better audio and picture quality, and there are more options on digital broadcasting

Explanation:

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3 years ago
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Answer:

Refracted

Explanation:

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