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Lyrx [107]
3 years ago
13

A thin coil has 17 rectangular turns of wire. When a current of 4 A runs through the coil, there is a total flux of 5 ✕ 10−3 T ·

m2 enclosed by one turn of the coil (note that
Φ = kI,

and you can calculate the proportionality constant k). Determine the inductance in henries.
Physics
1 answer:
Alex73 [517]3 years ago
7 0

Answer:

Inductance, L = 0.0212 Henries

Explanation:

It is given that,

Number of turns, N = 17

Current through the coil, I = 4 A

The total flux enclosed by the one turn of the coil, \phi=5\times 10^{-3}\ Tm^2

The relation between the self inductance and the magnetic flux is given by :

L=\dfrac{N\phi}{I}

L=\dfrac{17\times 5\times 10^{-3}}{4}

L = 0.0212 Henries

So, the inductance of the coil is 0.0212 Henries. Hence, this is the required solution.

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Answer the following questions for a mass that is hanging on a spring and oscillating up and down with simple harmonic motion. N
LiRa [457]

Answer:

1. equilibrium

2. bottom

3. bottom

4. nowhere

5. bottom

6. top & bottom

7. equilibrium

8. equilibrium

1. No

2. Yes

Explanation:

According to the following equation of motion for SHM:

x(t) = A\cos(\omega t + \phi)

where A is the amplitude, ω is the angular frequency, and ∅ is the phase angle.

Furthermore, the velocity and acceleration functions are as follows:

y(t) = -\omega A\sin(\omega t + \phi)\\a(t) = -\omega^2 A\cos(\omega t + \phi)

1. The acceleration is zero at the equilibrium. At the equilibrium, the net force on the object is zero. And according to Newton's Second Law, if the net force is zero, then the acceleration is zero as well.

2. The forces on the object in a vertical spring are the weight of the object and the spring force.

F = mg - kx

Since mg is constant along the motion, then the net force is maximum at the amplitude. For the special case in this question, the mass is always below the rest length of the spring. So the net force is maximum at the lower amplitude, because x is greater in magnitude at the lower amplitude.  According to Newton's Second Law, acceleration is proportional to the net force, hence the acceleration is at a maximum at the bottom.

3. As explained above, the magnitude of the net force is at a maximum at the lower amplitude, that is bottom.

4. The spring force is defined by Hooke's Law: F = -kx. Since the oscillation is small enough so that the mass is always below the rest length of the spring, then x is always greater than zero, hence nowhere in the motion will the spring force becomes zero.

5. As explained above, the force of gravity is constant and the spring force is proportional to the displacement, x. Therefore, the spring force is at a maximum at the lower amplitude, that is bottom.

6. The speed is zero when the mass is instantaneously at rest, that is the amplitude.

7. The net force on the mass is zero at the equilibrium.

8. The speed is at a maximum at the equilibrium.

1.  We will use the equation of motions given above. For simplicity, let's take ∅ = 0. At half its amplitude:

\frac{A}{2} = A\cos(\omega t)\\\frac{1}{2} = \cos(\omega t)\\\omega t = \pi / 3

Then the velocity at that point is

v(t) = -\omega A\sin(\pi /3) = -\omega A (0.866)

The maximum speed is where the acceleration is equal to zero:

0 = -\omega^2 A\cos(\omega t)\\\omega t = \pi / 2\\v_{max} = -\omega A\sin(\pi /2) = -\omega A

Comparing the maximum velocity to the velocity at A/2 yields that it is not half the maximum velocity:

-\omega A(0.866) \neq -\omega A

2. The maximum acceleration is at the amplitude.

A = A\cos(\omega t)\\\omega t = 2\pi\\a_{max} = -\omega^2 A\cos(2\pi) = -\omega^2 A

And the acceleration at A/2 is

\frac{A}{2} = A\cos(\omega t)\\\omega t = \pi / 3\\a(t) = -\omega^2 A\cos(\pi / 3) = -\omega^2 A (0.5)

Comparing these two results yields that the acceleration at half the amplitude is half the maximum acceleration.

5 0
3 years ago
I honestly don't know the answer to this.
Art [367]
The debates the picture was giving you a hint 


'
The debates the picture was giving you a hint " TIME " and what got finished?? The debates.













8 0
4 years ago
A pulley in the shape of a solid cylinder of mass 1.50 kg and radius 0.240 m is free to rotate around a horizontal shaft along t
liq [111]

Answer:

the speed of the textbook just before it hits the floor is 2.4 m/s

Explanation:

  Given the data in the question;

mass of pulley = 1.50 kg

radius of pulley = 0.240 m

mass of text book = 2.0 kg

height from which text book was released = 0.9 m

angular speed of the pulley = 10.0 rad/s

the speed of the textbook just before it hits the floor = ?

the speed of the textbook v = angular speed of the pulley × radius of pulley

we substitute

v = 10.0 rad/s × 0.240 m

v = 2.4 m/s

Therefore, the speed of the textbook just before it hits the floor is 2.4 m/s

6 0
3 years ago
What are some possible results of constructive forces on earth's surface
Vitek1552 [10]

Answer:

what are the choices

Explanation:

4 0
3 years ago
Read 2 more answers
Could you help me please !!!
grandymaker [24]

Answer:

6500g

Explanation:

7000 - 500 = 6500

7 0
3 years ago
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