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PtichkaEL [24]
3 years ago
10

In an electromagnetics lab, you are studying two coils which have a mutual inductance of M=300 mH. Suppose that the current in t

he 1st coil increased linearly from 2.8 A to 10 A in a span of 300 ms and that the 2nd coil has a resistance of R=0.4 Ohm, what is the magnitude of the induced current I in the 2nd coil? (Note: write only the two digits in the space provided for value of current in amperes).
Physics
1 answer:
Firdavs [7]3 years ago
7 0

Answer:

Explanation:

Given that,

The mutual inductance of the two coils is

M = 300mH = 300 × 10^-3 H

M = 0.3 H

Current increase in the coil from 2.8A to 10A

∆I = I_2 - I_1 = 10 - 2.8

∆I = 7.2 A

Within the time 300ms

t = 300ms = 300 × 10^-3

t = 0.3s

Second Coil resistance

R_2 = 0.4 ohms

We want to find the current in the second coil,

The same induced EMF is in both coils, so let find the EMF,

From faradays law

ε = Mdi/dt

ε = M•∆I / ∆t

ε = 0.3 × 7.2 / 0.3

ε = 7.2 Volts

Now, this is the voltage across both coils,

Applying ohms law to the second coil, V=IR

ε = I_2•R_2

0.72 = I_2 • 0.4

I_2 = 0.72 / 0.4

I_2 = 1.8 Amps

The current in the second coil is 1.8A

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agasfer [191]

v = 2.45×10^3\:\text{m/s}

Explanation:

Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as

F = \dfrac{d{p}}{d{t}} (1)

Assuming that the velocity remains constant then

F = \dfrac{d}{dt}(mv) = v\dfrac{dm}{dt}

Solving for v, we get

v = \dfrac{F}{\left(\frac{dm}{dt}\right)}\;\;\;\;\;\;\;(2)

Before we plug in the given values, we need to convert them first to their appropriate units:

The thrust <em>F</em><em> </em> is

F = 7.5×10^6\:\text{lbs}×\dfrac{4.45\:\text{N}}{1\:\text{lb}} = 3.34×10^7\:\text{N}

The exhaust rate dm/dt is

\dfrac{dm}{dt} = 15\dfrac{T}{s}×\dfrac{2000\:\text{lbs}}{1\:\text{T}}×\dfrac{1\:\text{kg}}{2.2\:\text{lbs}}

\;\;\;\;\;= 1.36×10^4\:\text{kg/s}

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v = \dfrac{F}{\left(\frac{dm}{dt}\right)} = \dfrac{3.34×10^7\:\text{N}}{1.36×10^4\:\text{kg/s}}

\;\;\;= 2.45×10^3\:\text{m/s}

6 0
2 years ago
This is for physical science. Can someone help me?
Radda [10]
A. 320 g
B. 160 g
C. 80 g
D. 40 g
6 0
3 years ago
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Problem 8 I estimate that the Gauss gun (a solenoid) is wound with 500 turns over a distance of 15cm with an average radius of 1
stellarik [79]

Answer:

Energy stored, U = 66.6 J

Explanation:

It is given that,

Number of turns in the solenoid, n = 500

Radius of solenoid, r = 1.5 cm = 0.015 m

Distance, d = 15 cm = 0.15 m

Let U is the energy stored in the solenoid. Its formula is given by :

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

L is the self inductance of the solenoid

L=\mu_o N^2A d

N is the no of turns per unit length

L=\mu_o (n/d)^2A d

L=\dfrac{4\pi\cdot10^{-7}\cdot500^{2}\cdot\pi\cdot\left(0.015\right)^{2}}{0.15}

L = 0.00148 Henry

U=\dfrac{1}{2}\times 0.00148\times 300^2

U = 66.6 J

Out of given options, the correct option for the energy stored in the solenoid is 70 J. So, the correct option is (a) "70 J".  

6 0
3 years ago
What are the major features of the human eye and to what are they analogous in a camera?
adelina 88 [10]

Answer:

The human major features with the analogous of camera are mentioned.

Explanation:

There are the following major features of human eye which are analogue with the camera:

  • Camera will have shutters which controls light entering into it, human eye consists of Diaphragm which also functions same.
  •  Both gives Real & inverted images.
  •  In camera film records image, in eye image is focused on retina and it is converted into electrical impulses.
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8 0
3 years ago
To check this out, you estimate that the induced voltage across a wild goose's wingtips to be approximately 7.54E-4 volts at lev
Klio2033 [76]

Answer:

1.45m

Explanation:

Detailed explanation and calculation is shown in the image below

6 0
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