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Contact [7]
3 years ago
14

A solenoidal inductor for a printed circuit board is being redesigned. To save weight, the number of turns is reduced by one-fif

th, with the geometric dimensions kept the same. By how much must the current change if the energy stored in the inductor is to remain the same? Select one: a. It must be two times larger. b. It should be one-half as large. c. It should be left the same. d. It must be four times larger. e. No change in the current can compensate for the reduction in the number of turns.
Physics
1 answer:
AlladinOne [14]3 years ago
8 0

The "it must be five times larger" current change if the energy stored in the inductor is to remain the same.

<u>Explanation:</u>

A current produced by a modifying magnetic field in a conductor is proportional to the magnetic field change rate named INDUCTANCE (L). The expression for the Energy Stored, that equation is given by:

U= \frac{1}{2} LI^2

Here L is the inductance and I is the current.

Here, energy stored (U) is proportional to the number of turns (N) and the current (I).

L = \frac{\mu_0 N^2 *A}{l}

mu not - permeability of core material

A -area of cross section

l - length

N - no. of turns in solenoid inductor

Now,given that the proportion always remains same:

\frac{N_2}{N_1} = \frac{I_1}{I_2}

In this way the expression

\frac{1}{5} = \frac{I_1}{I_2}

I_2 = I_1 \times 5

Thus, it suggest that "it must be five times larger" current change if the energy stored in the inductor is to remain the same.

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A ball rolls horizontally off a table and a height of 1.4 m with a speed of 4 m/s. How long does it take the ball to reach the g
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Given values:

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Plug in these values to get H(t):

H(t) = 1.4 + 0t - 4.905t²

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We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:

1.4 - 4.905t² = 0

4.905t² = 1.4

t² = 0.2854

t = ±0.5342s

Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))

t = 0.53s

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A physics student of mass 43.0 kg is standing at the edge of the flat roof of a building, 12.0 m above the sidewalk. An unfriend
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Answer:

The speed of the student just before she lands, v₂ is approximately 8.225 m/s

Explanation:

The given parameters are;

The mass of the physic student, m = 43.0 kg

The height at which the student is standing, h = 12.0 m

The radius of the wheel, r = 0.300 m

The moment of inertia of the wheel, I = 9.60 kg·m²

The initial potential energy of the female student, P.E.₁ = m·g·h₁

Where;

m = 43.0 kg

g = The acceleration due to gravity ≈ 9.81 m/s²

h = 12.0 h

∴ P.E.₁ = 43 kg × 9.81 m/s² × 12.0 m = 5061.96 J

The kinetic rotational energy of the wheel and kinetic energy of the student supporting herself from the rope she grabs and steps off the roof, K₁, is given as follows;

K_1 = \dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2

The initial kinetic energy, 1/2·m·v₁² and the initial kinetic rotational energy, 1/2·m·ω₁² are 0

∴ K₁ = 0 + 0 = 0

The final potential energy of the student when lands. P.E.₂ = m·g·h₂ = 0

Where;

h₂ = 0 m

The final kinetic energy, K₂, of the wheel and student is give as follows;

K_2 = \dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2

Where;

v₂ = The speed of the student just before she lands

ω₂ = The angular velocity of the wheel just before she lands

By the conservation of energy, we have;

P.E.₁ + K₁ = P.E.₂ + K₂

∴ m·g·h₁ + \dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2 = m·g·h₂ + \dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2

Where;

ω₂ = v₂/r

∴ 5061.96 J + 0 = 0 + \dfrac{1}{2} \times 43.0 \, kg \times v_{2}^2+\dfrac{1}{2} \times 9.60 \, kg\cdot m^2 \cdot \left (\dfrac{v_2}{0.300 \, m} }\right ) ^2

5,061.96 J = 21.5 kg × v₂² + 53.\overline 3 kg × v₂² = 21.5 kg × v₂² + 160/3 kg × v₂²

v₂² = 5,061.96 J/(21.5 kg + 160/3 kg) ≈ 67.643118 m²/s²

v₂ ≈ √(67.643118 m²/s²) ≈ 8.22454363 m/s

The speed of the student just before she lands, v₂ ≈ 8.225 m/s.

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3 years ago
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