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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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To increase the acceleration due to gravity, we will place the new station in a shorter radius which is 1.2 RE.

<h3>What is acceleration due to gravity?</h3>
  • This is free fall of an object under the influence of gravitational pull.

The acceleration due to gravity of the new station due to Earth's gravity is calculated as follows;

g = \frac{GM}{R^2}

where;

  • g is the acceleration due to gravity
  • G is gravitational constant
  • M is the mass of the earth
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To increase the acceleration due to gravity, the radius of the earth must be decreased. Thus, we will place the new station in a shorter radius which is 1.2 RE.

Learn more about acceleration due to gravity here: brainly.com/question/88039

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3 years ago
A free-falling golf ball strikes the ground and exerts a force on it. Which sentences are true about this situation? A golf ball
Harlamova29_29 [7]

Answer:

The ground exerts an equal force on the golf ball

Explanation:

Third's Newton Law states that:

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In this problem, object A is the golf ball while object B is the ground, so we can say that:

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8 0
3 years ago
Read 2 more answers
Question 1 (1 point)
puteri [66]

Answer:

250 N

Explanation:

a =  \frac{vf - vi}{t}

Vf=final velocity

Vi =initial velocity

=70-20/2=25m/s^2

F=ma

=10kg * 25m/s^2

=250N

3 0
3 years ago
Please help! Will mark Brainliest.
valentinak56 [21]

Answer:

18 Nm

Explanation:

if the correct answer

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An offshore oil well is 2 kilometers off the coast. The refinery is 4 kilometers down the coast. Laying pipe in the ocean is twi
shusha [124]

Answer:

Rectangular path

Solution:

As per the question:

Length, a = 4 km

Height, h = 2 km

In order to minimize the cost let us denote the side of the square bottom be 'a'

Thus the area of the bottom of the square, A = a^{2}

Let the height of the bin be 'h'

Therefore the total area, A_{t} = 4ah

The cost is:

C = 2sh

Volume of the box, V = a^{2}h = 4^{2}\times 2 = 128            (1)

Total cost, C_{t} = 2a^{2} + 2ah            (2)

From eqn (1):

h = \frac{128}{a^{2}}

Using the above value in eqn (1):

C(a) = 2a^{2} + 2a\frac{128}{a^{2}} = 2a^{2} + \frac{256}{a}

C(a) = 2a^{2} + \frac{256}{a}

Differentiating the above eqn w.r.t 'a':

C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}

For the required solution equating the above eqn to zero:

\frac{4a^{3} - 256}{a^{2}} = 0

\frac{4a^{3} - 256}{a^{2}} = 0

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Also

h = \frac{128}{4^{2}} = 8

The path in order to minimize the cost must be a rectangle.

8 0
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