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3 years ago

Which represents a type of electromagnetic wave? A. slinky wave B. jump rope wave C. shortwave radio waves D. ocean wave

Physics

2 answers:

Dahasolnce [82]3 years ago

8 0

The only item on that list that's an electromagnetic wave
is C). shortwave radio waves.

earnstyle [38]3 years ago

7 0

The answer should be (C) short wave radio waves.

Electromagnetic waves are transverse waves, similar to shortwave radio waves.

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A toroid having a square cross section, 5.49 cm on a side, and an inner radius of 18.1 cm has 450 turns and carries a current of

joja [24]

Answer:

(a) 4.27 x 10^-4 Telsa

(b) 3.28 x 10^-4 Telsa

Explanation:

side of square, a = 5.49 cm

inner radius, r = 18.1 cm = 0.181 m

number of turns,N = 450

current, i = 0.859 A

(a)

The magnetic field due to a solenoid due to inner radius is

$B = \frac{\mu_{0}Ni}{2\pi r_{inner}}$

$B = \frac{4\pi\times 10^{-7}\times 450\times 0.859}{2\pi \times 0.181}$

B = 4.27 x 10^-4 Telsa

(b)

The outer radius is R = 18.1 + 5.49 = 23.59 cm = 0.236 m

The magnetic field due to the outer radius is

$B = \frac{\mu_{0}Ni}{2\pi r_{outer}}$

$B = \frac{4\pi\times 10^{-7}\times 450\times 0.859}{2\pi \times 0.236}$

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3 years ago

A horizontal beam of light of intensity 25 W/m2 is sent through two polarizing sheets. The polarizing direction of the first mak

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Answer:

option (B)

Explanation:

Intensity of unpolarised light, I = 25 W/m^2

When it passes from first polarisr, the intensity of light becomes

$I'=\frac{I_{0}}{2}=\frac{25}{2}=12.5 W/m^{2}$

Let the intensity of light as it passes from second polariser is I''.

According to the law of Malus

$I'' = I' Cos^{2}\theta$

Where, θ be the angle between the axis first polariser and the second polariser.

$I'' = 12.5\times Cos^{2}15$

I'' = 11.66 W/m^2

I'' = 11.7 W/m^2

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3 years ago

A point charge of 9.00 × 10−9 C is located at the origin of a coordinate system. A positive charge of 3.00 × 10−9 C is brought i

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A point charge is located at the origin of a coordinate system. A positive charge is brought in from infinity to a point. The charges are at distance for given electrical potential energy is 3.34 x 10⁷ m.

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The electric potential energy is the work done by a test charge to bring it from infinity to a particular location.

The electric potential energy is given by the relation,

V = kQ/r

where k = 9 x 10⁹ J.m/C ,Q = 3 x 10⁻⁹ C, V =8.09 × 10⁻⁷ J.

Substitute the values into the expression to get the distance between the charges.

8.09 × 10⁻⁷ = 9 x 10⁹ x 3 x 10⁻⁹ / r

r =3.34 x 10⁷ m

Thus, the distance between the charges will be 3.34 x 10⁷ m.

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