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bixtya [17]
3 years ago
10

How do electric and magnetic fields interact to create electromagnets?​

Physics
1 answer:
professor190 [17]3 years ago
3 0

Answer:

8548458

Explanation:

95895

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the answer is that the dough has the same mass before and after it was flattened

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What is the principal of moment​
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hope it helps...

Explanation:

The Principle of Moments states that when a body is balanced, the total clockwise moment about a point equals the total anticlockwise moment about the same point.

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slader A jet is circling an airport control tower at a distance of 15.9 km. An observer in the tower watches the jet cross in fr
liubo4ka [24]

Answer:

y = 138.96 m

Explanation:

The angle subtended by the moon is the mean of the angle of the arc between the two most extreme points of the moon, we can see that the angle is very small, so we can approximate this arc to a straight line and then use the trigonometric relationships

         sin θ = y / L

where L = 15.9 10³ m and θ = 8.74 10⁻³ rad

          y = L sin θ

          y = 15.9 10³ sin (8.74 10⁻³)

         y = 15.9 10³    0.0087399

         y = 138.96 m

4 0
3 years ago
Please help!! I’m in a rush :(
Burka [1]
Kinetic energy is a form of energy that an object or particle has by reaction of its motion.
3 0
3 years ago
A long cylindrical insulating shell has an inner radius of a = 1.41 m and an outer radius of b = 1.67 m. The shell has a constan
Natasha2012 [34]

Answer:

a. E = 122.4 N/C

b. E = 58.2 N/C

c. E = 0

Explanation:

The electric field at an arbitrary point away from the axis of the cylinder can found by applying Gauss’ Law, which states that an electric flux through a closed surface is equal to the total charge enclosed by this surface divided by electric permittivity.

In order to apply this law, we have to draw an imaginary cylindrical surface of arbitrary height ‘h’ and radius ‘r’, which is equal to the point where the E-field is asked.

A. For the outside of the cylinder, we will draw our imaginary surface with r = 1.97.

E2\pi rh = \frac{\lambda V}{\epsilon_0} = \frac{\lambda \pi (b^2 - a^2)h}{\epsilon_0}\\E2\pi (1.97)h = \frac{(5.3\times 10^{-9})\pi(1.67^2 - 1.41^2)h}{\epsilon_0}\\E = 122.4~N/C

B. This time our imaginary surface should be inside the cylinder, therefore the enclosed charge will be less than that of part A.

E2\pi rh = \frac{\lambda V_{enc}}{\epsilon_0} = \frac{\lambda \pi (r^2 - a^2}h{\epsilon_0}\\E2\pi (1.51)h = \frac{5.3\times 10^{-9})\pi(1.51^2 - 1.41^2)h}{\epsilon_0}\\E = 58.2~N/C

C. In this case our imaginary surface will be inside the cylinder, where there is no charge at all. Therefore, the enclosed charge will be zero and the electric field will be zero.

6 0
3 years ago
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