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Alika [10]
3 years ago
9

Induced EMF and Current in a Shrinking LoopShrinking Loop. A circular loop of flexible iron wire has an initial circumference of

161cm , but its circumference is decreasing at a constant rate of 13.0cm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 1.00T , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop.Find the magnitude of the emf EMF induced in the loop after exactly time 8.00s has passed since the circumference of the loop started to decrease.
Physics
1 answer:
vodomira [7]3 years ago
4 0

Answer:

Explanation:

Let c be the circumference and r be the radius

c = 2πr , r = c / 2π , area A = π r² = π (c/2π )²  = (1/4π) x c²

flux (ψ) = BA = 1 X 1/4π X c²

dψ/dt = 1/4π x 2c dc/dt =1/2π x c x dc/dt

at t = 8 s

c = 161 - 13 x 8 = 57 cm , dc/dt = 13 cm/s  

e = dψ/dt = (1 / 2π )x 57 x 13 x 10⁻⁴ = 118 x 10⁻⁴ V.

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Two vehicles A and B accelerate uniformly from rest.
spayn [35]

Answer:

(i) Please find attached the required velocity time graphs plotted with MS Excel

(ii) The velocity of vehicle A at the 18th second = 20 m/s

The velocity of vehicle B at the 18th second = 0 m/s

(iii) The distance between the two vehicles at the moment in (ii) above is 60 meters

Explanation:

The given parameters of the motion of vehicles A and B are;

The acceleration of vehicles A and B = Uniform acceleration starting from rest

The maximum velocity attained by vehicle A = 30 m/s

The time it takes vehicle A to attain maximum velocity = 10 s

The maximum velocity attained by vehicle B = 30 m/s

The time it takes vehicle B to attain maximum velocity = The time it takes vehicle A to attain maximum velocity = 10 s

The time duration vehicle A maintains its maximum velocity = 6 s

The time duration vehicle B maintains its maximum velocity = 4 s

(i) From the question, we get the following table;

\begin{array}{ccc}Time &V_A&V_B\\0&0&0\\10&30&40\\14&30&40\\16&30&20\\18&20&0\\22&0&\end{array}

From the above table the velocity time graphs of vehicles A and B is created with MS Excel and can included here

(ii) The velocity of vehicle A at the start = 0 m/s

After accelerating for 10 seconds, the velocity of vehicle A = The maximum velocity of vehicle A = 30 m/s

The maximum velocity is maintained for 6 seconds which gives;

At 10 s + 6 s = 16 s, the velocity of vehicle A = 30 m/s

The time it takes vehicle A to decelerate to rest = 6 s

The deceleration of vehicle A, a_A = (30 m/s - 0 m/s)/(6 s) = 5 m/s²

Therefore, we get;

v = u - a_A·t

At the 18th second, the deceleration time, t = 18 s - 16 s = 2 s

u = 30 m/s

∴ v₁₈ = 30 - 5 × 2 = 20

The velocity of vehicle A at the 18th second, V_{18A} = 20 m/s

For vehicle B, we have;

At the 14th second, the velocity of vehicle B = 40 m/s

Vehicle B decelerates to rest in, t = 4 s

The deceleration of vehicle B, a_B = (40 m/s - 0 m/s)/(4 s) = 10 m/s²

For vehicle B, at the 18th second, t = 18 s - 14 s = 4 s

∴ v_{18B} = 40 m/s - 10 m/s² × 4 s = 0 m/s

The velocity of the vehicle B at 18th second, v_{18B} = 0 m/s

(iii) The distance covered by vehicle A up to the 18th second is given by the area under the velocity-time graph as follows;

The area triangle A₁ = (1/2) × 10 × 30 = 150

Area of rectangle, A₂ = 6 × 30 = 180

Area of trapezoid, A₃ = (1/2) × (30 + 20) × 2 = 50

The distance covered in the 18th second by vehicle S_A = A₁ + A₂ + A₃

∴ S_A = 150 + 180 + 50 = 380

The distance covered in the 18th second by vehicle S_A = 380 m

The distance covered by the vehicle B in the 18th second is given by the area under the velocity time graph of vehicle B as follows;

Area of trapezoid, A₅ = (1/2) × (18 + 4) × 40 = 440

The distance covered by the trapezoid, S_B = 440 m

The distance of the two vehicles apart at the 18t second, S_{AB} = S_B - S_A

∴ S_{AB} = 440 m - 380 m = 60 m

The distance of the two vehicles from one another at the 18th second, S_{AB} = 60 m.

5 0
3 years ago
The form of energy given off by the vibrating strings of the violin is? A. Electrical Energy B. Potential Energy C. Radiant Ener
S_A_V [24]

Answer:

D . Sound energy

Explanation:

When the strings of a violin vibrate it produces sound which is sound energy. Due to the vibration of the strings the air present near the strings also vibrate in resonance with the strings. This compreesion and decompression's produced in the air is nothing but the sound. So the form of energy given off by the vibrating strings of the violin is Sound energy.

6 0
3 years ago
A spectrograph helps astronomers to determine:
Bezzdna [24]

Answer:

I may be wrong sir/ma’am, but I believe it’s 1. Surface temperatures. 3.radio signals from space. And 4. Distance of stars.

Explanation:

sorry y’all:(

8 0
2 years ago
The density is 1.5 g/cm^3. the mass is 1500. what is the volume
Naddik [55]
The density of an object can be calculated using the formula Density = Mass/Volume. In this case however we are searching for the volume and must rearrange the formula so that we are solving for the volume. If you multiply both sides by volume and then divide both sides by mass you end up with the equation Volume = Mass/Density.

Volume = 1500g/1.5g/cm^3
Volume = 1000 cm^3
8 0
3 years ago
A moving object has a kinetic energy of 150 J and a momentum of 25.1 kg·m/s.
PolarNik [594]

Answer:

11.95m/s

Explanation:

A moving object has a kinetic energy of 150 J and a momentum of 25.1 kg·m/s.

a) Find the speed of the object. Answer in units of m

K. E =½mv²

150= ½mv²

Multiply both sides by 2

mv² = 300

Divide both sides by v²

m = 300/v² ..................  Equation 1

Momentum is the product of mass and velocity

Momentum = mv

25.1 = mv

Divide both sides by v

m = 25.1/v ................ Equation 2

Equate equations 1 and 2

300/v² = 25.1/v

Cross multiply

25.1v² = 300v

Multiply v with both sides

25.1v = 300

Divide both sides by 25.1

v = 300/25.1

V = 11.95m/s

I hope this was helpful, please mark as brainliest

8 0
3 years ago
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