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3 years ago

11

What is the relationship between the density of the equipotential lines, the density of the electric field lines and the strengt

h of the electric field?

1 answer:

12345 [234]3 years ago

7 0

Answer:

I dont. understand the question, maybe insert the picture?

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Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The

IRISSAK [1]

To develop this problem we will apply the concepts related to the Electromagnetic Force. The magnetic force can be defined as the product between the free space constant, the current (of each cable) and the length of these, on the perimeter of the cross section, in this case circular. Mathematically it can be expressed as,

$F= \frac{\mu_0}{2\pi} \frac{I^2L}{d}$

Here,

$\mu_0$ = Permeability free space

I = Current

L = Length

d= Distance between them

Our values are,

$L = 0.67m$

$m = 0.082kg$

$d = 7.4*10^{-3}m$

$I = \text{Current through each of the wires}$

Rearranging the previous equation to find the current,

$\frac{mg}{L} = 2*10^{-7} (\frac{I^2}{d})$

$I = \sqrt{\frac{mgd}{(2*10^{-7})L}}$

$I = \sqrt{\frac{(0.082)(9.8)(7.4*10^{-3})}{(2*10^{-7})(0.67)}}$

$I = \sqrt{44377.9}$

$I = 210.66A$

Therefore the current in the rods is 210.6A

8 0

4 years ago

Read 2 more answers

An object moving due to gravity can be described by the motion equation y=y0+v0t−12gt2, where t is time, y is the height at tha

Lynna [10]

Answer: 6.45 s

Explanation:

We have the following equation:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0$ is the height when the rock hits the ground

$y_{o}=75 m$ the height at the edge of the cilff

$V_{o}=20 m/s$ the initial velocity

$g=9.8 m/s^{2}$ acceleration due gravity

$t$ time

$0=75 m+(20 m/s)t-(4.9 m/s^{2})t^{2}$ (2)

Rearranging the equation:

$-(4.9 m/s^{2})t^{2} + (20 m/s)t + 75 m=0$ (3)

At this point we have a quadratic equation of the form $at^{2}+bt+c=0$ , and we have to use the quadratic formula if we want to find $t$ :

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ (4)

Where $a=-4.9$ , $b=20$ , $c=75$

Substituting the known values and choosing the positive result of the equation:

$t=\frac{-20\pm\sqrt{20^{2}-4(-4.9)(75)}}{2(-4.9)}$ (5)

$t=6.453 s$ This is the time it takes to the rock to hit the ground

8 0

3 years ago

How is wind direction indicated on a weather map ?

kotegsom [21]

It uses arrows to show what way the wind is going

6 0

3 years ago

During the deceleration of an ascending elevator, the normal force on the feet of a passenger is _____ her weight. During the de

gizmo_the_mogwai [7]

Answer: Smaller than ; larger than

Explanation:

When the elevator is moving in the upward direction, then the force acting on it is negative in nature because of

N= mg +ma, (g is gravity and a is acceleration)

here ma is negative so the N= mg-ma

Hence, it feels smaller than its original weight.

When the elevator is moving downward , then the force acting will be positive in nature

N= mg+ma,

here ma will be positive so it feels larger the original weight of passenger.

7 0

3 years ago

The water that powers the generators enters and leaves the system at a low speed (thus we can neglect its change of kinetic ener

madam [21]

Answer:

Explanation:

Let density of water be ρ .

During flow , volume of water flowing per second is constant

loss of P. E per unit volume = ρ gh , 83.5 % is lost

Gain of K E per unit volume = 1/2 ρ v²

83.5 % of mgh = ρ 1/2 ρ v²

1/2 ρ v² = .835 x 9.8

v² = 2 x .835 x 9.8

= 16.366

v = 4.04 m /s

4 0

4 years ago