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EleoNora [17]
3 years ago
8

What type of force pulls in two opposite directions?

Physics
2 answers:
Natasha2012 [34]3 years ago
7 0
Magnets facing the same way  <span />
Elanso [62]3 years ago
7 0

tension (associated with normal faults)

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If the voltage across a circuit of constant resistance is doubled, the power dissipated by that circuit will
densk [106]

Answer:

The voltage will quadruple

Explanation:

The power dissipated in a circuit is given by

P=\frac{V^2}{R}

where

V is the voltage

R is the resistance

In this problem, the voltage across the circuit is doubled:

V' = 2V

So the new power dissipated is

P'=\frac{V'^2}{R}=\frac{(2V)^2}{R}=4\frac{V^2}{R}=4 P

so, the power dissipated will quadruple.

6 0
3 years ago
Using Gauss's law, calculate the electric field at a point distance s from a long wire bearing uniform charge density. i need he
11111nata11111 [884]

Answer:

E = 2k  \frac{\lambda}{ r}

Explanation:

Gauss's law states that the electric flux equals the wax charge between the dielectric permeability.

We must define a Gaussian surface that takes advantage of the symmetry of the problem, let's use a cylinder with the faces perpendicular to the line of charge. Therefore the angle between the cylinder side area has the same direction of the electric field which is radial.

            Ф = ∫ E . dA = E ∫ dA = q_{int} /ε₀

tells us that the linear charge density is

            λ = q_ {int} /l

            q_ {int} = l λ

we substitute

            E A = l λ /ε₀

is area of ​​cylinder is

           A = 2π r l

we substitute

            E = \frac{ l \ \lambda}{ \epsilon_o \ 2\pi  \ r \ l }

             E = \frac{\lambda}{ 2\pi  \epsilon_o \ r}

the amount

            k = 1 / 4πε₀

            E = 2k  \frac{\lambda}{ r}

5 0
2 years ago
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gfhvgtrtjrgvfjrrgfrfftuyrisnhdvfcgfridkjhsybvvtfvjfcgvwjfccegvghcvgrcgvrekgvrkgvkvvrvkvfgkerruuyti

4 0
2 years ago
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An 85.0-N crate of apples sits at rest on a ramp that runs from the ground to the bed of a truck. The ramp is inclined at 28.0°
irina [24]

Answer:

A  75.1 N and a direction of 152° to the vertical.

B 85.0 N at 0° to the vertical.

Explanation:

A) The interaction partner of this normal force has what magnitude and direction?

The interaction partner of this normal force is the component of the weight of the crate perpendicular to the ramp. <u>It has a magnitude of 85cos28° = 75.1 N and a direction of 180° - 28° = 152° to the vertical(since it is directed downwards perpendicular to the ramp).</u>

B) The normal and frictional forces are perpendicular components of the contact force exerted on the crate by the ramp. What is the magnitude and direction of the contact force?

Since this force has to balance the weight of the crate, its magnitude is 85.0 N. Its direction has to be vertically opposite to that of the weight.

Since the weight is 180° to the vertical (since it is directed downwards), this force is 0° to the vertical.

<u>So, this force has a magnitude of 85.0 N and a direction of 0° to the vertical.</u>

8 0
2 years ago
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