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andrew-mc [135]
2 years ago
8

A square conducting loop 8.4 cm on a side is placed in a uniform B-field so that the plane of the loop is perpendicular to the d

irection of the field lines. If the loop is then converted into a rectangular loop measuring 2.1 cm on its shortest side in 6.50 ms, and the average emf induced across the loop is 14.7 V during this time period, what isA) the strength of the B-field?B) What is the direction of the induced current in the conducting loop (assume the B-field is directed out of the page, show direction in your picture or with the words "Clockwise" or "Counter-Clockwise")?
Physics
1 answer:
arsen [322]2 years ago
4 0

Answer:

Explanation:

area of square loop A = side²

= 8.4² x 10⁻⁴

A = 70.56 x 10⁻⁴ m²

when it is converted into rectangle , length = 14.7  , width = 2.1

area = length x width

= 14.7 x 2.1 x 10⁻⁴

= 30.87 x 10⁻⁴ m²

Let magnetic field be B

Change in flux = magnetic field x change in area

= B x ( 70.56 x 10⁻⁴ - 30.87 x 10⁻⁴ )

= 39.69 x 10⁻⁴ B

rate of change of flux = change in flux / time taken

= 39.69 x 10⁻⁴ B  / 6.5 x 10⁻³

= 6.1 x 10⁻¹ B

emf induced = 6.1 x 10⁻¹ B

6.1 x 10⁻¹ B  = 14.7 ( given )

B = 2.41 x 10

= 24.1 T

B ) magnetic flux is decreasing , so it needs to be increased as per Lenz's law . Hence current induced will be anticlockwise so that additional  magnetic flux is induced out of the page.

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Answer:

the friction force on this box is closest to 45.9 N.

Explanation:

given data

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coefficient of static friction = 0.600

solution

we know box does not move relative to the wagon

we get here friction force  that is express as

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here mass = weight ÷ g

mass = \frac{150}{9.8} = 15.3 kg

put value in equation 1 we get

friction force = 15.3 kg × 3

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3 0
2 years ago
A particle traveling in a circular path of radius 300 m has an instantaneous velocity of 30 m/s and its velocity is increasing a
grigory [225]
<h2>Answer:</h2>

(c) 5m/s²

<h2>Explanation:</h2>

Total acceleration (a) of a particle in a circular motion is the vector sum of the radial or centripetal acceleration (a_{C}) of the particle and the tangential acceleration (a_{T}) of the particle and its magnitude can be calculated as follows;

a = \sqrt{(a_{C})^2 + (a_{T})^2}           ---------------------(i)

<em>But;</em>

a_{C} = \frac{v^{2} }{r}      ------------------------------(ii)

Where;

v = instantaneous velocity

r =  radius of the circular path of motion

<em>From the question;</em>

v = 30m/s

r = 300m

(i) First let's calculate the centripetal acceleration by substituting the values above into equation (ii) as follows;

a_{C} = \frac{30^{2} }{300}

a_{C} = \frac{900}{300}

a_{C} = 3m/s²

(ii) From the question, the velocity of the particle is increasing at a constant rate of 4m/s²  and that is the tangential acceleration a_{T}, of the particle. i.e;

a_{T} = 4m/s²

(iii) Now substitute the values of a_{C} and a_{T} into equation (i) as follows;

a = \sqrt{(3)^2 + (4)^2}

a = \sqrt{(9) + (16)}

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5 0
2 years ago
A box is sliding with a speed of 4.50 m/s on a horizontal surface when, at point P, it encounters a rough section. The coefficie
Rudiy27

Answer:

a) xf = 5.1 m

b) u = 0.304

c) x = 10.3 m

Explanation:

we will use the following formula:

u = 0.1 + A*x

Si x = 12.5 m, u = 0.6

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Clearing x:

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jenyasd209 [6]

Answer:

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7 0
2 years ago
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