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yan [13]
3 years ago
9

The magnetic field due to a utility wire is 0.10 mT when you are at a distance of 10 meters from it. What current (in Amperes) f

lows through the wire?
Physics
1 answer:
ale4655 [162]3 years ago
4 0

Answer:

I = 5000 A

Explanation:

We will use Ampere's Law to calculate the current:

B = \frac{\mu I}{2\pi r}\\\\

where,

B = Magnetic Field Strength = 0.1 mT = 1 x 10⁻⁴ T

μ = Permeability of Free Space = 4π x 10⁻⁷ N/A²

I = Current = ?

r = radius = 10 m

Therefore,

1\ x\ 10^{-4}\ T = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(I)}{2\pi(10\ m)}\\\\I = \frac{(1\ x\ 10^{-4}\ T)(2\pi (10\ m))}{4\pi\ x\ 10^{-7}\ N/A^2}

<u>I = 5000 A</u>

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Novay_Z [31]

This question is incomplete, the complete question is;

A weightlifter holds a 1,300 N barbell 1 meter above the ground. One end of a 2-meter-long chain hangs from the center of the barbell. The chain has a total weight of 400 N. How much work (in J) is required to lift the barbell to a height of 2 m?

What is the average force (average with respect to height of the barbell from the ground) exerted by the weightlifter in the process?

Answer: Average force exerted by the weightlifter in the process = 1600N

Explanation:

To find Work done to lift a barbell and half of the hanging chain we say;

W₁ = ( 1300N + (1/2 × 400N)) × 1m

W₁ = (1300 + 200) Nm

W₁ = 1500J

now work done to lift the upper half of the chain we say:

W₂ = (1/2 × 400N) ×  (1/2 × 1m)

W₂ = 200N × 0.5m

W₂ = 100J

So total work done will be

W = W₁ + W₂

W = 1500J + 100J

W = 1600J

To find the average force exerted by the weight lifter, we say;

F = W/D

F = (1600 / 1m) N

F = 1600N

∴Average force = 1600N

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3 years ago
If a cart of 2 kg mass has a force of 8 newtons exerted on it, what is its acceleration?
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Force = mass * acceleration

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8 N =  2 kg * a

8 = 2a

2a = 8

a = 8/2 = 4

acceleration = 4 m/s²
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Answer:

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The "unstretched length" is shorter than the actual unloaded length because the load applied to the material caused the extension of the material thereby increasing its length.

<h3>Hooke's law</h3>

Hooke's law states that the force applied to an elastic material is directly proportional to the extension of the elastic material.

F = kx

where;

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Thus, we can conclude that, the "unstretched length" is shorter than the actual unloaded length because the load applied to the material caused extension of the material thereby increasing the length of the material.

Learn more about Hooke's law here: brainly.com/question/2648431

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