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4 years ago

What is the relationship between the strength of an

Physics

1 answer:

Anastaziya [24]4 years ago

3 0

Answer:

As the number of turns in the coil increases, the strength of the electromagnet increases.

Explanation:

When current flows through a coil the coil behaves as an electromagnet. The strength of electromagnet depend the amount of current, no of turns of coil and the core of coil.

B=μ₀ N I

μ₀ = permeability of the core

N = Number of turns of the coil

I = Current flowing through the coil

Increasing the current and number of coils increase the strength of electromagnet.

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A car has weight of 10,ooow What is the mass of the car on earth (g=10N/KG)

Gre4nikov [31]

Answer:

1000 kg

Explanation:

We know the formula for finding out weight is W=mg

W= 10,000 N

M=?

g=10 N/KG

W=mg

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3 years ago

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4 years ago

Suppose an asteroid orbiting the sun had an orbital period of 7. 5 years. What would its orbital radius be?.

creativ13 [48]

By using the orbital period equation we will find that the orbital radius is r = 4.29*10^11 m

<h3>What is the orbital period?</h3>

This would be the time that a given body does a complete revolution in its orbit.

It can be written as:

$T = \sqrt{\frac{4*\pi ^2*r^3}{G*M} }$

Where:

π = 3.14
G is the gravitational constant = 6.67*10^(-11) m^3/(kg*s^2)
M is the mass of the sun = 1.989*10^30 kg
r is the radius, which we want to find.

Rewriting the equation for the radius we get:

$T = \sqrt{\frac{4*\pi ^2*r^3}{G*M} }\\\\r = \sqrt[3]{ \frac{T^2*G*M}{4*\pi ^2} }$

Where T = 7.5 years = 7.5*(3.154*10^7 s) = 2.3655*10^8 s

Replacing the values in the equation we get:

$r = \sqrt[3]{ \frac{(2.3655*10^8 s)^2*(6.67*10^{-11} m^3/(kg*s^2))*(1.989*10^{30} kg)}{4*3.14 ^2} } = 4.29*10^{11 }m$

So the orbital radius is 4.29*10^11 m

If you want to learn more about orbits, you can read:

brainly.com/question/11996385

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2 years ago

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4 years ago

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A 12 N force is used to pull a wagon 10 m to the left in 15 seconds. Which of the following represents the magnitude of the forc

Lana71 [14]

1. - 12N

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3 years ago

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