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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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Katyanochek1 [597]
That would be 0 degrees Celsius aka the melting point of water.... If you look at the diagram I attached you notice that at 0 degrees Celsius it is flat, this is because much heat is needed at this point for water to rise to 1 degree... It is the same for the boiling point (100)

Hope this helps! (If correct, please rank as brainliest answer) :)
7 0
3 years ago
Question 8 a-e plz
N76 [4]

Answer:

(a) t = 0 s

(b) t = 0 s, 30 s, 55 s

(c) t = 40 s to t = 60 s

(d) t = 10 s to t = 15 s

(e) a = 6 m/s^2

Explanation:

(a) The car is at starting position at t = 0 s and v = 0 m/s.

(b) The velocity of car is zero when the time is t = 0 s, 30 s and 55 s.

(c) from t = 40 s to 60 s the car is moving in the negative direction.

(d) The fastest speed is 60m/s from t = 10 s to t = 15 s.

(e) The slope of the velocity time graph gives acceleration.

a = (60 - 0) / (10 - 0) = 6 m/s^2

5 0
3 years ago
Calculate the electric field intensity at a point 3 cm away from point charge of 3 x 10^-9 C.
lukranit [14]

Answer:

The electric field intensity is <u>30000 N/C.</u>

Explanation:

Given:

Magnitude of the point charge is, q=3\times 10^{-9}\ C

Distance of the given point from the point charge is, d=3\ cm=0.03\ m

Electric field intensity is directly proportional to the magnitude of point charge and inversely proportional to the square of the distance of the point and the given charge.

Therefore, electric field intensity 'E' at a distance of 'd' from a point charge 'q' is given as:

E=\frac{kq}{d^2}

Plug in k=9\times 10^9\ N\cdot m^2/C^2, q=3\times 10^{-9}\ C, d=0.03\ m. Solve for 'E'.

E=\frac{(9\times 10^9\ N\cdot m^2/C^2)(3\times 10^{-9}\ C)}{(0.03\ m)^2}\\\\E=\frac{27}{0.0009}\ N/C\\\\E=30000\ N/C

Therefore, the electric field intensity at a point 3 cm from the point charge is 30000 N/C.

3 0
3 years ago
An electric scooter has a battery capable of supplying 135 Wh of energy. If friction forces and other losses account for 60.0% o
Vedmedyk [2.9K]

Answer:

206 m

Explanation:

The energy output of the scooter battery = 135 Wh = 135 W × 3600 s = 486000 J.

If 60.0% accounts for losses, then

useful energy = (100 - 60)% of 486000 J = 40% × 486000 J = 194400 J

The energy needed to move through an altitude = product of weight and altitude

E=F\times h

h = \dfrac{E}{F} = \dfrac{194400\text{ J}}{945\text{ N}} = 206\text{ m}

7 0
3 years ago
A 2.0kg pendulum swings from point A of a height of 0.10m to a point B of the same height. The heights are relative to the lower
mr_godi [17]
Please press “Thanks!” after viewing my answer:
The change in Gravitational Potential Energy (GPE) is actually 0! Think about it:
What is the formula for it?
Well, GPE = mgh.
So, when the height is the same in both cases, we know the gravitational potential energy is the same. Therefore, change in GPE = 0.
Sure, GPE did transform to KE during the swing, but all of that KE transformed back into the original amount of GPE.

Hope this helped! Please press “Thanks!”
4 0
3 years ago
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