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podryga [215]
3 years ago
7

What is the peak emf generated by rotating a a980-turn, 11cm diameter coil in the Earth’s 5·10−5 T magnetic field, given the p

lane of the coil is originally perpendicular to the Earth’s field and is rotated to be parallel to the field in 7 ms? g
Physics
2 answers:
Andreyy893 years ago
7 0

Answer:

E = 426mV

Explanation:

Given N = 980

B = 5×10-⁵ T

R = 11cm = 0.11m

Δt = 7ms = 7×10-³s

Area A = π×D²/4 = π×0.11²/4 = 0.0095m²

E = - N×ΔBAωSinθ = NBA×2π/t

The field and the Area are not changing but the angle between the area vector changes from 0 to 90° and Sinθ changes from 0 to 1. Therefore the flux changes from zero to maximum.

So E = - 980×5.10×10-⁵×0.0095×2π/(7×10-³)

E = 426×10-³V= 426mV

andrezito [222]3 years ago
4 0

Answer:

The peak emf generated by the coil is 418.3 mV

Explanation:

Given;

number of turns, N = 980 turns

diameter, d = 11 cm = 0.11 m

magnetic field, B = 5 x 10⁻⁵ T

time, t = 7 ms = 7 x 10⁻³ s

peak emf, V₀ = ?

V₀ = NABω

Where;

N is the number of turns

A is the area

B is the magnetic field strength

ω is the angular velocity

V₀ = NABω and ω = 2πf = 2π/t

V₀ = NAB2π/t

A = πd²/4

V₀ = N x (πd²/4) x B x (2π/t)

V₀ = 980 x (π x 0.11²/4) x 5 x 10⁻⁵ x (2π/0.007)

V₀ = 980 x 0.00951 x  5 x 10⁻⁵ x 897.71

V₀ = 0.4183 V = 418.3 mV

The peak emf generated by the coil is 418.3 mV

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Answer:

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Explanation:

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I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

4 0
3 years ago
If you ride your bike 20 miles and it takes you 120 minutes,what is your average speed
Alexus [3.1K]
120 minutes=2 hours
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5 0
3 years ago
A rock is thrown at a window that is located 16.0 m above the ground. The rock is thrown from the ground at an angle of 40.0° ab
Angelina_Jolie [31]

Answer:

x = 27.3 m

Explanation:

This is a projectile launching exercise, let's start by looking for the time it takes for the rock to reach the height of the window.

Let's use trigonometry to find the velocities of the rock

           sin 40 = v_{oy} / v

           cos 40 = v₀ₓ / v

           v_{oy}= v sin 40

           v₀ₓ = v cos 40

           v_{oy} = 30 sin 40 = 19.28 m / s

           v₀ₓ = v cos 40

           v₀ₓ = 30 cos 40 = 22.98 m / s

we look for the time

        v_{y}^2 = v_{oy}^2 - 2 g y

         v_{y}^2 = 19.28 2 - 2 9.8 16 = 371.71 - 313.6 = 58.118

        v_{y} = 7.623 m / s

we calculate the time

          v_{y} = v_{oy} - gt

          t = (v_{oy} - v_{y}) / g

          t = (19.28 -7.623) / 9.8

          t = 1,189 s

           

since the time is the same for both movements let's use this time to find the horizontal distance

           x = v₀ₓ t

           x = 22.98 1,189

           x = 27.3 m

4 0
4 years ago
a chuck wagon with an initial velocity of 4 m/s and a mass of 35 kg gets a push with 350 joules of force. what is the wagon's fi
Kitty [74]

Answer:

the final velocity of the wagon is 6 m/s.

Explanation:

Given;

initial velocity of the wagon, u = 4 m/s

mass of the wagon, m = 35 kg

energy applied to the wagon, E = 350 J

The final velocity of the wagon is calculated as;

E = ¹/₂m(v² - u²)

m(v^2-u^2) = 2E\\\\v^2-u^2 = \frac{2E}{m} \\\\v^2 =  \frac{2E}{m}  + u^2\\\\v = \sqrt{\frac{2E}{m}  + u^2} \\\\v = \sqrt{\frac{2(350)}{35}  + (4)^2}\\\\v = 6 \ m/s

Therefore, the final velocity of the wagon is 6 m/s.

8 0
3 years ago
Someone please help me i don't know the answer to this one
Andreas93 [3]

Answer:

it is the bottom one

Explanation:

the wavelength is shorter but more compressed.

3 0
3 years ago
Read 2 more answers
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