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asambeis [7]
3 years ago
13

Suppose that a rectangular toroid has 2,000 windings and a self-inductance of 0.060 H. If the height of the rectangular toroid i

s h = 0.60 m, what is the current (in A) flowing through the rectangular toroid when the energy in its magnetic field is 4.0 ✕ 10−6 J?
Physics
1 answer:
andrey2020 [161]3 years ago
3 0

Answer:

0.01154 A

Explanation:

We have given the energy in the magnetic field U=4\times 10^{-6}J

Value of inductance L =0.060 H

Energy stored in magnetic field is given by U=\frac{1}{2}Li^2

i=\sqrt{\frac{2U}{L}}

i=\sqrt{\frac{2\times 4\times 10^{-6}}{0.06}}=0.01154\ A

So the current flowing through rectangular toroid will be 0.01154 A

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Answer:

B

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Electronegativity is<br>____ at the top of the periodic table.​
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Answer:

Electronegativity is a chemical property at the top of the periodic table

Explanation:

electronegativity is a chemical property that measures the ability of an atom to attract a bonding pair of electrons when it is part of a compound

it's a property of atom that increases as you go to the right and up.

One example is the Pauling scale. it is a numerical scale of electronegativities. It was first developed by Linus Pauling.

it is commonly used to calculate the ability of electronegativity to attract electrons to itself.

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3 0
4 years ago
Read 2 more answers
n a car lift used in a service station, compressed air exerts a force on a small piston that has a circular cross section of rad
Oksi-84 [34.3K]

Answer: 1477.78 N

Explanation:

Let's assume that the cross sectional area of the smaller piston be A1

let's also assume the cross sectional area of the larger piston be A2

We assume the force applied to the smaller piston be F1

We also assume the force applied to the larger piston be F2

we then use the formula

F1/A1 = F2/A2

From our question,

The radius of the smaller piston is 5 cm = 0.05 m

The radius of the larger piston is 15 cm = 0.15 m

The force of the larger piston is 13300 N

The force of the smaller piston is unknown = F

A1 = πr² = 3.142 * 0.05² = 0.007855 m²

A2 = πr² = 3.142 * 0.15² = 0.070695 m²

F1/0.007855 = 13300/0.070695

F1 = (13300 * 0.007855) / 0.070695

F1 = 104.4715 / 0.070695

F1 = 1477.78 N

Thus, the force the compressed air must exert is 1477.78 N

5 0
3 years ago
Butter is an example for<br>liquid in solid or solid in liquid ​
ioda

Answer:

Liquid in solid.

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Butter is colloidal in nature belonging to the category of gels. This means that the dispersed phase is liquid and the dispersing medium is solid. In short, butter is an example for liquid in solid.

8 0
3 years ago
1 2 3 4 5 6 7 8 9 10
7nadin3 [17]

Answer:

<h2>C. <u>0.55 m/s towards the right</u></h2>

Explanation:

Using the conservation of law of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.

Momentum = Mass (M) * Velocity(V)

BEFORE COLLISION

Momentum of 0.25kg body moving at 1.0m/s = 0.25*1 = 0.25kgm/s

Momentum of 0.15kg body moving at 0.0m/s(body at rest) = 0kgm/s

AFTER COLLISION

Momentum of 0.25kg body moving at x m/s = 0.25* x= 0.25x kgm/s

<u>x is the final velocity of the 0.25kg ball</u>

Momentum of 0.15kg body moving at 0.75m/s(body at rest) =

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Using the law of conservation of momentum;

0.25+0 = 0.25x + 0.1125

0.25x = 0.25-0.1125

0.25x = 0.1375

x = 0.1375/0.25

x = 0.55m/s

Since the 0.15 kg ball moves off to the right after collision, the 0.25 kg ball will move at <u>0.55 m/s towards the right</u>

<u></u>

4 0
3 years ago
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