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alexdok [17]
3 years ago
11

What is the electric field 3.3 m from the center of the terminal of a Van de Graaff with a 7.20 mC charge, noting that the field

is equivalent to that of a point charge at the center of the terminal?
Physics
1 answer:
Snowcat [4.5K]3 years ago
6 0

Answer:

Electric Field is 5.943801*10^6 N/C

Explanation:

Electric Field:

It originates from positive charge and ends at negative charge.

General Formula for electric Field:

E=\frac{kq}{r^2}

where:

k is the Coulomb Constant

q is the charge

r is the distance

Given:

q=7.20 mC

r=3.3 meters

k=8.99*10^9 N.m^2/C^2

Find:

Electric Field=?

Solution:

E=\frac{kq}{r^2}

E=\frac{(8.99*10^9)(7.20*10^{-3})}{3.3^2}\\E=5943801.653 N/C\\E=5.943801653*10^6 N/C

Electric Field is 5.943801*10^6 N/C

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aleksandr82 [10.1K]

Answer:

d = 4 d₀o

Explanation:

We can solve this exercise using the relationship between work and the variation of kinetic energy

         W = ΔK

In that case as the car stops v_f = 0

the work is

          W = -fr d

we substitute

          - fr d₀ = 0 - ½ m v₀²

           d₀ = ½ m v₀² / fr

now they indicate that the vehicle is coming at twice the speed

          v = 2 v₀

using the same expressions we find

           d = ½ m (2v₀)² / fr

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3 0
3 years ago
Learning Goal: To understand the behavior ofthe electric field at the surface of a conductor, and itsrelationship to surface cha
Ivan

Complete Question

The complete question is shown on the first uploaded image  

Answer:

a   it is always zero

b  0

c  \eta  =  -\epsilon _o E

Explanation:ss

Here the  net charge is  on the outer surface of the conductor thus this means that the net charge inside the conductor is zero

Generally the charge density of a conductor is dependent on the charge per unit area  which implies that the charge density is dependent on the net charge  so this  means that the charge density inside the conductor is zero

 

Generally the direction of electric field this from the  positive charge to the negative charge  so from the question we can deduce  that the negative charge is located on the surface of the conductor

    So We can mathematically define the charge density on the surface of the electric field as

             ∮E \cdot dA =  \frac{-Q}{\epsilon _o}

Where E is the electric field

          dA change in unit area

           -Q is the negative charge

          \epsilon _o  is the permittivity of free space

So

          EA  =  \frac{-Q}{\epsilon _o }

           \frac{Q}{A}  =  -\epsilon _o E

          \eta  =  -\epsilon _o E

Where \eta is the charge density

   

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3 years ago
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Answer:

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Never gonna let you down

Never gonna run around and desert you

Haha Rick rolled you

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3 0
3 years ago
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A fuel pump sends gasoline from a car's fuel tank to the engine at a rate of 5.64 10-2 kg/s. the density of the gasoline is 735
Irina18 [472]
Given:
Gasoline pumping rate, R = 5.64 x 10⁻² kg/s
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Let v =  speed of pumping the gasoline, m/s
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The gasoline pumping rate is given as 5.64 x 10⁻² kg/s, therefore
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Answer:  2.076 m/s
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Certain neutron stars (extremely dense stars) are believed to be rotating at about 10 rev/s. If such a star has a radius of 18 k
Aleks [24]

Answer:

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Explanation:

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\frac{GM_{ns}}{R^2}= \omega^2 R

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3 years ago
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