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zavuch27 [327]
3 years ago
6

A (20*20) cm² loop has a resistance of 0.10 Ω. A magnetic field perpendicular to the loop is B = 4t - 2t², where B is in tesla a

nd t is in seconds.
What is the current in the loop at t = 0.0 s, t = 1.0 s, and t = 2.0 s?
Physics
2 answers:
Ilya [14]3 years ago
8 0

Answer with Explanation:

We are given that

Area of loop=(20\times 20) cm^2=400\times 10^{-4} m^2

1 cm^2=10^{-4} m^2

Resistance, R=0.1\Omega

B=4t-2t^2

We know that magnetic flux

\phi=BA

Emf ,E=\mid \frac{d\phi}{dt}\mid =\mid\frac{d(BA}{dt}\mid =\mid A\frac{dB}{dt}=400\times 10^{-4}\times \frac{4t-2t^2}{dt}\mid =\mid400\times 10^{-4}\times(4-4t)\mid

Current, I=\frac{E}{R}

Current, I=\frac{\mid 400\times 10^{-4}(4-4t)\mid }{0.1}=1.6\mid (1-t)\mid

Substitute t=0 s

Then, I=1.6\mid (1-0)\mid=1.6 A

Substitute t=1 s

Then, I=1.6\mid (1-1)\mid=0

Substitute

t=2 s

Current, I=1.6\mid(1-2)\mid=1.6 A

Sliva [168]3 years ago
8 0

Answer:

Explanation:

Area, A  = 20 x 20 cm² = 0.04 m²

Resistance, R = 0.10 ohm

Magnetic field, B = 4t - 2t²

the induced emf is given by

e=\frac{d\phi }{dt}

where, Ф i the flux linked with the coil.

e=A\times \frac{dB }{dt}

e=0.04(4 - 4t)

Induced current

i = e/ R

i=\frac{0.04(4 - 4t)}{0.1}

i = 0.4(4 - 4t)

For t = 0 s

i = 0.4 x 4 = 1.6 A

For t = 1 s

i = 0.4 ( 4 - 4) = 0 A

For t = 2 s

i = 0.4 ( 4 - 8) = 1.6 A in opposite direction

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3 years ago
Find the difference between two masses measured as 123.6 grams and 115.972 grams. Express the answer to the correct number of si
xxTIMURxx [149]

Answer:

The difference is 7.6 grams.

Explanation:

In mathematics the difference of two numbers is express as the subtraction between them:

         

a-b

So to find out the difference between the two measured masses, a will be represented by 123.6 grams since is the bigger number, and b by 115.972 grams.

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8 0
3 years ago
A real object is 10.0 cm to the left of a thin, diverging lens having a focal length of magnitude 16.0 cm. What is the location
amm1812

Answer:

A)6.15 cm to the left of the lens

Explanation:

We can solve the problem by using the lens equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where

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f is the focal length

p is the distance of the object from the lens

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f=-16.0 cm (the focal length is negative for a diverging lens)

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Solvign the equation for q, we find

\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}

q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm

And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is

A)6.15 cm to the left of the lens

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