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Angelina_Jolie [31]
3 years ago
15

At what frequency should a 200-turn, flat coil of cross sectional area of 300 cm2 be rotated in a uniform 30-mT magnetic field t

o have a maximum value of the induced emf equal to 8.0 V
Physics
1 answer:
stellarik [79]3 years ago
5 0

Answer:

The frequency of the coil is 7.07 Hz

Explanation:

Given;

number of turn of the coil, N = 200 turn

area of the coil, A = 300 cm² = 0.03 m²

magnitude of magnetic field, B = 30 mT = 0.03 T

maximum value of induced emf, E = 8 V

The maximum induced emf in the coil is given by;

E = NBAω

E = NBA(2πf)

f = \frac{E_{max}}{2\pi*NBA}

where;

f is the frequency of the coil

f = \frac{E_{max}}{2\pi*NBA}\\\\f = \frac{8}{2\pi(200)(0.03)(0.03)} \\\\f = 7.07 \ Hz

Therefore, the frequency of the coil is 7.07 Hz

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Sophie [7]

Answer:

<u>Principal</u><u> </u><u>focus</u><u> </u><u>of</u><u> </u><u>concav</u><u>e</u><u> </u><u>lens</u><u> </u><u>-</u><u> </u>

★ The point at which rays parallel to principal axis coming from infinity appear to converge after being refracted from concave lens is called the principal focus of concave lens.

<em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em>

• <u>Additional</u><u> information</u><u> </u><u>-</u><u> </u>

★ Principal focus - A number of rays parallel to the principal axis after reflection from a concave mirror meet at a point on the principal axis or appear to come from a point after reflection from a convex mirror on the principal axis. This is called principal focus.

8 0
3 years ago
1 1.1.2 Quiz: The Nature of Physics
Natasha2012 [34]

Answer: Theory (D.)

Explanation:

A theory is based on years of careful scientific study, observation, and experimentation.

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3 0
4 years ago
Kinetic and
beks73 [17]

Answer:

raise the board to a higher angle

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Static friction is dependent on the angle of inclination, it means as the angle of incline increases, the force of friction will increases as normal force will decrease.

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Hence, the correct option is "raise the board to a higher angle".

5 0
3 years ago
What happened to the weight of an object when it is taken from Earth to the Moon? why?<br>​
Sholpan [36]

Answer:

the weight of the object decreases when it is taken from the Earth to the Moon

Explanation:

The weight of an object is defined as the product of the mass of the object with the acceleration due to gravity of the Planet.

W =mg

where,

W = weight of the object

m = mass of the object

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The mass of an object remains constant everywhere in the universe. Therefore, the weight is directly proportional to the value of acceleration due to gravity.

The value of acceleration due to gravity on the Moon is lesser than its value on the Earth.

<u>Hence, the weight of the object decreases when it is taken from the Earth to the Moon </u>

6 0
3 years ago
The large blade of a helicopter is rotating in a horizontal circle. The length of the blade is 7.17 m, measured from its tip to
gayaneshka [121]

Answer:

1.789

Explanation:

let the tangential velocity at the tip of the blade (7.17m from center) be V₁ and the tangential velocity at 4.01 be V₂.

recall that tangential velocity can be related to angular velocity, ω by the following relationship:

V = rω, where V is the tangential velocity, ω is angular velocity and r = radius.

rearranging, we get

ω = V/r

We also know that at any point in the rotation, even though the tangential velocity at 7.17m radius (V₁)   will be different from the tangential velocity at 4.01m radius  (V₂), their angular velocity will be the same, hence we can equate:

ω₁ = ω₂, or

V₁/r₁ = V₂/r₂ (rearranging)

V₁/V₂ = r₁/r₂ -----(eq 1)

We also know that centripetal acceleration can be expressed in terms of tangential velocity and radius, i.e

a =V²/r

to find the ratio of the centripetal acceleration at the tip and at r=4.01m

a₁/a₂ = (V₁²/r₁)   / ( V₂² / r₂)   (rearranging & simplify)

=  (V₁/V₂)² (r₂ / r₁)   (substituting eq1 into equation)

=  (r₁/r₂)²(r₂ / r₁)  (simplifying)

= (r₁/r₂)    (substituting r₁=7.17 and r₂=4.01)

= 7.17 / 4.01

= 1.789

7 0
4 years ago
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