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slava [35]
3 years ago
7

A long line carrying a uniform linear charge density+50.0μC/m runs parallel to and 10.0cm from the surface of a large, flat plas

tic sheet that has a uniform surface charge density of −100μC/m2on one side.Part AFind the location of all points where an α particle would feel no force due to this arrangement of charged objects.L=_____________ m from the line.Part BChoose an appropriate location of these points at a distance calculated in part A.o above the lineo between the line and the sheet.
Physics
1 answer:
IRISSAK [1]3 years ago
6 0

Answer:

The location of the points is at the distance  above the line charge Option A.

Explanation:

Part A

The electric field due to a line of charge is  expressed as :

Eₐ = λ /2πε₀L  …… (1)

while  the electric field due to a non-conducting elastic sheet is expressed as:

Еₙ =   λ /2ε₀  …… (2)

In order for the  particle to experience no force, the electric field at that point is zero.

Equating  (1) and (2) together, we have:.

λ /2πε₀L = σ /2ε₀  

Rearranging the above expression, we have:

L =  λ /πσ

Substituting 50.0 μC/m  for λ, 3.14 for  π  , and 100 μC/ m² for σ, we have:

L=  50.0 μC/m / (3.14)* 100 μC/ m²

L = 0.159m

From the L value obtained above, the location of all points where an  particle would experience no force due to the arrangement of charged objects is L 0.159m  from the line.

Part B  

The electric field is zero at all the points above the line with the distance 0.159m  from the line.

Thus, the location of the points is at the distance  above the line charge Option A.

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Answer:

n = 1.42

Explanation:

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n=\dfrac{c}{v}\\\\n=\dfrac{3\times 10^8}{2.1\times 10^8}\\\\n = 1.42

So, the refractive index of the medium is 1.42.

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3 years ago
Combine these three velocity vectors into a resultant: 3.0 m/s north, 4.0 m/s east 1.0 m/s west. Identify the resultant vector
Slav-nsk [51]

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Explanation:

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5 0
3 years ago
A fluid flows through a pipe whose cross-sectional area changes from 2.00 m2 to 0.50 m2 . If the fluid’s speed in the wide part
borishaifa [10]

Answer:

v₂ = 7/ (0.5)= 14 m/s

Explanation:

Flow rate of the fluid

Flow rate is the amount of fluid that circulates through a section of the pipeline (pipe, pipeline, river, canal, ...) per unit of time.

The formula for calculated the flow rate is:

Q= v*A Formula (1)

Where :

Q is the Flow rate (m³/s)

A is the cross sectional area of a section of the pipe (m²)

v is the speed of the fluid in that section (m/s)

Equation of continuity

The volume flow rate Q for an incompressible fluid at any point along a pipe is the same as the volume flow rate at any other point along a pipe:

Q₁= Q₂

Data

A₁ = 2m² : cross sectional area 1

v₁ = 3.5 m/s : fluid speed through A₁

A₂ = 0.5 m² : cross sectional area 2

Calculation of the fluid speed through A₂

We aply the equation of continuity:

Q₁= Q₂

We aply the equation of Formula (1):

v₁*A₁= v₂*A₂

We replace data

(3.5)*(2)= v₂*(0.5)

7 = v₂*(0.5)

v₂ = 7/ (0.5)

v₂ =  14 m/s

4 0
3 years ago
The static frictional force between a 95-kilogram object and the floor is 45 Newtons. The kinetic frictional force is only 22 Ne
Lisa [10]

Answer:

F = 69.5 [N]

Explanation:

We must remember that the friction force is defined as the product of the normal force by the coefficient of friction, and it can be calculated by the following expression.

f=N*miu

where:

N = normal force [N]

miu = friction coefficient

f = friction force = 22 [N]

Now we must calculate the force exerted by means of Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

F - f = m*a

where:

F = force exerted [N]

f = friction force [N]

m = mass = 95 [kg]

a = acceleration = 0.5 [m/s²]

Now replacing:

F - 22 = 95*0.5\\F = 47.5 + 22\\F = 69.5 [N]

6 0
3 years ago
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