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Setler [38]
3 years ago
10

A 100-turn, 3.0-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 60%u2218 away from vertical increase

s from 0.50 T to 2.5T in 0.90s .
What is the induced emf in the coil?
Physics
1 answer:
Xelga [282]3 years ago
8 0

Answer:

Induced emf in the coil, E = 0.157 volts

Explanation:

It is given that,

Number of turns, N = 100

Diameter of the coil, d = 3 cm = 0.03 m

Radius of the coil, r = 0.015 m

A uniform magnetic field increases from 0.5 T to 2.5 T in 0.9 s.

Due to this change in magnetic field, an emf is induced in the coil which is given by :

E=-NA\dfrac{\Delta B}{\Delta t}

E=-100\times \pi (0.015)^2\times \dfrac{2.5-0.5}{0.9}

E = -0.157 volts

Minus sign shows the direction of induced emf in the coil. Hence, the induced emf in the coil is 0.157 volts.

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No. The protostellar cloud spins faster in the collapsing stage (stage 1) and becomes much slower in the contraction stage (stage 2)

Explanation:

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Explanation:

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3 years ago
A ball is thrown horizontally from the top of a 60 m building and lands 100 m from the base of the building. How long is the bal
zhannawk [14.2K]

Answer:

The ball is in the air for 3.5 seconds

The initial horizontal component of velocity is 28.6 m/s

The vertical component of the final velocity is 34.3 m/s downward

The final velocity is 44.7 m/s in the direction 50.2° below the horizontal

Explanation:

A ball is thrown horizontally

That means the vertical component of the initial velocity u_{y}=0

The initial velocity is the horizontal component u_{x}

The ball is thrown from the top of a 60 m

That means the vertical displacement component y = 60 m

→ y = u_{y} t + \frac{1}{2} gt²

where g is the acceleration of gravity and t is the time

y = -60 m , g = -9.8 m/s² , u_{y}=0

Substitute these values in the rule

→ -60 = 0 + \frac{1}{2} (-9.8)t²

→ -60 = -4.9t²

Divide both sides by -4.9

→ 12.2449 = t²

Take √ for both sides

∴ t = 3.5 seconds

* <em>The ball is in the air for 3.5 seconds </em>

The initial velocity is the horizontal component u_{x}

The ball lands 100 meter from the base of the building

That means the horizontal displacement x = 100 m

→ x = u_{x} t

→ t = 3.5 s , x = 100 m

Substitute these values in the rule

→ 100 = u_{x} (3.5)

Divide both sides by 3.5

→ u_{x} = 28.57 m/s

<em>The initial horizontal component of velocity is 28.6 m/s</em>

The vertical component of the final velocity is v_{y}

→ v_{y} = u_{y} + gt

→ u_{y} = 0 , g = -9.8 m/s² , t = 3.5 s

Substitute these values in the rule

→ v_{y} = 0 + (-9.8)(3.5)

→ v_{y} = -34.3 m/s

<em>The vertical component of the final velocity is 34.3 m/s downward</em>

The final velocity v is the resultant vector of  v_{x} and v_{y}

→ Its magnetude is v=\sqrt{(v_{x})^{2}+(v_{y})^{2}}

→ Its direction tan^{-1}\frac{v_{y}}{v_{x}}

→ v_{y} = 28.6 , v_{y} = -34.3

Substitute this values in the rules above

→ v=\sqrt{(28.6)^{2}+(-34.3)^{2}}=44.66

→ Its direction tan^{-1}\frac{-34.3}{28.6}=-50.18

The negative sign means the direction is below the horizontal

<em>The final velocity is 44.7 m/s in the direction 50.2° below the horizontal</em>

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Explanation:

I hope this helps a little bit.

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