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motikmotik
3 years ago
6

When a coil is carrying a current of 25.0 A that is increasing at 145 A/s the induced emf in the coil has magnitude 3.70 mV.

Physics
1 answer:
Aleks04 [339]3 years ago
6 0

Answer:

a) 25.5 µH

b) 22.95 mV

Explanation:

Induced emf in a inductor is given by

E = L * di/dt, where

E is the voltage of the circuit

L is the inductance of the circuit

di/dt if the rate of inductance

A

So we have

0.0037 = L * 145

L = 0.0037 / 145

L = 0.0000255

L = 25.5 µH

B

i(t) = 225t²

Recall that

E = L * di/dt, so that

E = 25.5 µH * |225t²|

Differentiating with respect to t, we have

E = 25.5 * 2 * 225t

E = 25.5 * 450t

Solving for t = 2,we get

E = 25.5 * 450(2)

E = 25.5 * 900

E = 22950 µV or

E = 22.95 mV

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According to Bernoulli's principle force is given by

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Damage to the ___________ will most likely cause a person to lose the ability to comprehend language. (2 points)
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Write a hypothesis for Part II of the lab, which is about the relationship described by F = ma. In the lab, you will use a toy c
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2 years ago
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A spring-mass system has a spring constant of 3 Nm. A mass of 2 kg is attached to the spring, and the motion takes place in a vi
frosja888 [35]

Answer:

The answer to the question

The steady state response is u₂(t) = -\frac{3\sqrt{2} }{2}cos(3t + π/4)

of the form R·cos(ωt−δ) with R = -\frac{3\sqrt{2} }{2}, ω = 3 and δ = -π/4

Explanation:

To solve the question we note that the equation of motion is given by

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γ = Damping coefficient = 1

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Therefore we have 2·u'' + u' + 3·u = 27·cos(3·t)−18·sin(3·t)

The homogeneous equation 2·u'' + u' + 3·u is first solved as follows

2·u'' + u' + 3·u = 0 where putting the characteristic equation as

2·X² + X + 3 = 0 we have the solution given by \frac{-1+/-\sqrt{23} }{4} \sqrt{-1} =\frac{-1+/-\sqrt{23} }{4} i

This gives the general solution of the homogeneous equation as

u₁(t) = e^{(-1/4t)} (C_1cos(\frac{\sqrt{23} }{4}t) + C_2sin(\frac{\sqrt{23} }{4}t)

For a particular equation of the form 2·u''+u'+3·u = 27·cos(3·t)−18·sin(3·t) which is in the form u₂(t) = A·cos(3·t) + B·sin(3·t)

Then u₂'(t) = -3·A·sin(3·t) + 3·B·cos(3·t) also u₂''(t) = -9·A·cos(3·t) - 9·B·sin(3·t) from which  2·u₂''(t)+u₂'(t)+3·u₂(t) = (3·B-15·A)·cos(3·t) + (-3·A-15·B)·sin(3·t). Comparing with the equation 27·cos(3·t)−18·sin(3·t)  we have

3·B-15·A = 27

3·A +15·B = 18

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The general solution is then  u(t) = u₁(t) + u₂(t)

however since u₁(t) = e^{(-1/4t)} (C_1cos(\frac{\sqrt{23} }{4}t) + C_2sin(\frac{\sqrt{23} }{4}t) ⇒ 0 as t → ∞ the steady state response = u₂(t) = -\frac{3\sqrt{2} }{2}·cos(3·t + π/4) which is of the form R·cos(ωt−δ) where

R = -\frac{3\sqrt{2} }{2}

ω = 3 and

δ = -π/4

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3 years ago
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Nesterboy [21]

Answer:

The dog has more momentum than the pony.

Explanation:

To solve this problem, we must remember the formula for calculating momentum, which is given below:

momentum = p = m*v

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Using this knowledge, let's calculate the momentum for the dog and the pony.

Dog: p = m*v = (2kg)*(41 m/s) = 82 kg*m/s

Pony: p = m*v = (75kg)*(1 m/s) = 75 kg*m/s

Since 82 > 75, we can conclude that the dog has more momentum.

Hope this helps!

3 0
3 years ago
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