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Svetlanka [38]
2 years ago
5

During a 72-ms interval, a change in the current in a primary coil occurs. This change leads to the appearance of a 6.0-mA curre

nt in a nearby secondary coil. The secondary coil is part of a circuit in which the resistance is 12 Ω. The mutual inductance between the two coils is 3.2 mH. What is the change in the primary current?
Physics
1 answer:
Gwar [14]2 years ago
7 0

Answer:

so change in primary current is 1.620 A

Explanation:

Given data

current I =6.0-mA = 6 × 10^(-3) A

resistance R = 12 Ω

mutual inductance  M = 3.2 mH = 3.2 × 10^(-3) H

dt = 72-ms = 72 × 10^-(3) s

to find out

change in the primary current

solution

we know that

Electric and magnetic fields in secondary coil  = mutual inductance × change in primary current / dt      ............1

and we know also that Electric and magnetic fields in secondary coil = resistance × current

so = 6 × 10^(-3) × 12 = 72 × 10^(-3)  volts

so that we say

change in primary current from equation 1

change in primary current  = Electric and magnetic fields in secondary coil × dt / mutual inductance

change in primary current  = 72 × 10^(-3)  ×  72 × 10^(-3) / 3.2 × 10^(-3)

change in primary current  = 1.620

so change in primary current is 1.620 A

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A team of astronauts is on a mission to land on and explore a large asteroid. In addition to collecting samples and performing e
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<h2>Answer: 117.626m/s</h2>

Explanation:

The escape velocity V_{esc} is given by the following equation:

V_{esc}=\sqrt{\frac{2GM}{R}}   (1)

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On the other hand, we know the density of the asteroid is \rho=3.84(10)^{8}g/m^{3} and its volume is V=2.17(10)^{12}m^{3}.

The density of a body is given by:

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M=8.33(10)^{20}g=8.33(10)^{17}kg  (4)  This is the mass of the spherical asteroid

In addition, we know the volume of a sphere is given by the following formula:

V=\frac{4}{3}\piR^{3}   (5)

Finding R:

R=\sqrt[3]{\frac{3V}{4\pi}}   (6)

R=\sqrt[3]{\frac{3(2.17(10)^{12}m^{3})}{4\pi}}   (7)

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Now we have all the necessary elements to calculate the escape velocity from (1):

V_{esc}=\sqrt{\frac{2(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(8.33(10)^{17}kg)}{8031.38m}}   (9)

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5 0
2 years ago
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