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Lemur [1.5K]
3 years ago
6

Suppose that a rectangular toroid has 1,500 windings and a self-inductance of 0.060 H. If the height of the rectangular toroid i

s h = 0.80 m, what is the current (in A) flowing through the rectangular toroid when the energy in its magnetic field is 9.0 ✕ 10−6 J?
Physics
1 answer:
V125BC [204]3 years ago
5 0

Answer:

Current in the toroid will be I=17.32\times 10^{-3}A

Explanation:

We have given number of winding in rectangular toroid N = 1500

Self inductance of toroid L = 0.06 H

Magnetic energy stored in toroid E=9\times 10^{-16}J

We have to find the current in the toroid

Magnetic energy stored is equal to E=\frac{1}{2}Li^2

9\times 10^{-6}=\frac{1}{2}\times 0.06\times I^2

I=17.32\times 10^{-3}A

So current in the toroid will be I=17.32\times 10^{-3}A

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List down all the jovian planets in order of increasing distance from the sun
Olenka [21]

Answer:

There are total eight planets in the solar system and the average distance from the sun to each planet in increasing order is given below.

Explanation:

The average distance from the sun is listed below in increasing order.

1. Mercury  - It is the most closet planet to Sun, 57 million km

2. Venus  - 108 million km

3. Earth  - 150 million km

4. Mars  - 228 million km

5. Jupiter  - 779 million km

6. Saturn  - 1.43 billion km

7. Uranus  - 2.88 billion km

8. Neptun - It is the most farthest from the Sun, 4.50 billion km

8 0
3 years ago
If a net force is greater than 0 Newtons, what is the force?
goldenfox [79]

Answer:disequilibrium

Explanation:

When the net force is not zero it is called disequilibrium

5 0
3 years ago
Using calcium’s atomic structure, as shown in the image, what is this element’s atomic number?
ElenaW [278]

Answer:

Atomic number of calcium is 20.

6 0
3 years ago
What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally h
zysi [14]

Complete question:

What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)

Answer:

The peak emf generated by the coil is 15.721 kV

Explanation:

Given;

Radius of coil, r = 0.250 m

Number of turns, N = 500-turn

time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s

magnetic field strength, B = 0.425 T

Induced peak emf = NABω

where;

A is the area of the coil

A = πr²

ω is angular velocity

ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s =  60 rev/s

Induced peak emf = NABω

                               = 500 x (π x 0.25²) x 0.425 x 376.738

                               = 15721.16 V

                               = 15.721 kV

Therefore, the peak emf generated by the coil is 15.721 kV

5 0
3 years ago
A plane leaves Seattle, flies 76.0 mi at 22.0 ∘ north of east, and then changes direction to 51.0 ∘ south of east. After flying
mote1985 [20]
This can be solve by using a triangle, because the path of the plane formed a triangle. first solve the angle form by the second direction
angle = 180 - 51 - 22 = 107 degrees
then using the cosine law
c^2 = a^2 + b^2 - 2ab cos C
c^2 = 76^2 + 123^2 - 2 ( 76) ( 123) cos ( 107)
c = 162.4 mi <span>the crew fly to go directly to the field
</span>
5 0
3 years ago
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