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Fittoniya [83]
2 years ago
6

The north pole of a bar magnet points towards a thin circular coil of wire containing 40 turns. The magnet is moved away from th

e coil, so that the ux through one turn inside the coil decreases by ∆Φ = 0.3 T.m2 in a time ∆t = 0.2 s. What is the average EMF induced in the (whole) coil during this time interval? Viewed from the side opposite of the bar magnet (from the right), does the induced current run clockwise or counterclockwise? Explain briey. [2 p.]

Physics
1 answer:
givi [52]2 years ago
8 0

Answer:

<em>60 V</em>

<em>The direction will be anticlockwise from the right side of the magnet.</em>

Explanation:

The change in magnetic flux = ∆Φ = 0.3 T.m^2

The change in time = ∆t = 0.2 s

number of turns = 40 turns

The induced emf E = N∆Φ/∆t =

E = (40 x 0.3)/0.2 = <em>60 V</em>

If the magnet is moved away from the coil, the induced current on the coil will try to oppose the motion of the magnet by attracting the magnet towards the coil. For the magnet to be attracted towards the coil, it must possess the equivalent of a magnetic south pole. For the equivalent of a magnetic south pole, the current on the coil will flow in the clockwise direction when viewed from the left side of the magnet. <em>When viewed from the right side of the magnet, the direction will appear as anticlockwise.</em>

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Calculate the distance d from the center of the sun at which a particle experiences equal attractions from the earth and the sun
fredd [130]

Answer:

149.34 Giga meter is the distance d from the center of the sun at which a particle experiences equal attractions from the earth and the sun.

Explanation:

Mass of earth = m = 5.976\times 10^{24} kg

Mass of Sun = M = 333,000 m

Distance between Earth and Sun = r = 149.6 gm =  1.496\times 10^{11} m[/tex]

1 giga meter = 10^{9} meter

Let the mass of the particle be m' which x distance from Sun.

Distance of the particle from Earth = (r-x)

Force between Sun and particle:

F=G\frac{M\times m'}{x^2}=G\frac{333,000 m\times m'}{x^2}

Force between Sun and particle:

F'=G\frac{mm'}{(r-x)^2}

Force on particle is equal:

F = F'

G\frac{333,000 m\times m'}{x^2}=G\frac{mm'}{(r-x)^2}

\frac{x}{r-x}=\sqrt{333,000} = ±577.06

Case 1:

\frac{x}{r-x}=577.06

x = 1.49\times 10^{11} m=149.34 Gm

Acceptable as the particle will lie in between the straight line joining Earth and Sun.

Case 2:

\frac{x}{r-x}=-577.06

x = 1.49\times 10^{11} m=149.86 Gm

Not acceptable as the particle will lie beyond on line extending straight from the Earth and Sun.

3 0
2 years ago
The heaviest wild lion ever measured had a mass of 313 kg. Suppose this lion is walking by a lake when it sees an empty boat flo
11Alexandr11 [23.1K]

Answer:

The kinetic energy dissipated is 3286.5 J

Explanation:

K.E before collision = 1/2m1v1^2 = 1/2×313×6^2 = 5634 J

K.E after collision = 1/2(m1+m2)v2^2

From the law of conservation of momentum:

m1+m2 = m1v1/v2 = 313×6/2.5 = 751.2 kg

K.E after collision = 1/2×751.2×2.5^2 = 2347.5 J

K.E dissipated = 5634 J - 2347.5 J = 3286.5 J

3 0
2 years ago
a pulley of radius 0.9m is used to lift a bucket from the well. if it took 3.6 rotations for the pulley to take water out of the
Alex777 [14]

Answer:

h = 20.36[m]

Explanation:

To solve this problem we must calculate the perimeter (length of a circumference) of the circumference, which is denoted by the following equation:

L = 2*π*r

where:

r = radius = 0.9[m]

π = 3.1416

L = 2*(3.1416)*(0.9)

L = 5.654[m]

Now we know that the pulley or circumference had to rotate 3.6 times to get the water out of the well. In this way the depth of the well can be calculated by means of the following equation:

h = 3.6*L

h = 3.6*5.654

h = 20.36[m]

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Energy can change form, but the total amount of energy stays the same.
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2 years ago
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