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Ratling [72]
3 years ago
7

Plain electromagnetic wave (in air) has a frequency of 1 MHz and its B-field amplitude is 9 nT a. What is the wavelength in air?

b. What is the E-field amplitude? c. What is the intensity of this wave? What pressure does such wave exert on a totally-absorbing surface?
Physics
1 answer:
Norma-Jean [14]3 years ago
6 0

Answer:

Part a)

\lambda = 300 m

Part b)

E = 2.7 N/C

Part c)

I = 9.68 \times 10^{-3} W/m^2

P = 3.22 \times 10^{-11} N/m^2

Explanation:

Part a)

As we know that frequency = 1 MHz

speed of electromagnetic wave is same as speed of light

So the wavelength is given as

\lambda = \frac{c}{f}

\lambda = \frac{3\times 10^8}{1\times 10^6}

\lambda = 300 m

Part b)

As we know the relation between electric field and magnetic field

E = Bc

E = (9 \times 10^{-9})(3\times 10^8)

E = 2.7 N/C

Part c)

Intensity of wave is given as

I = \frac{1}{2}\epsilon_0E^2c

I = \frac{1}{2}(8.85 \times 10^{-12})(2.7)^2(3\times 10^8)

I = 9.68 \times 10^{-3} W/m^2

Pressure is defined as ratio of intensity and speed

P = \frac{I}{c} = \frac{9.68\times 10^{-3}}{3\times 10^8}

P = 3.22 \times 10^{-11} N/m^2

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For this problem, we use the Coulomb's law written in equation as:

F = kQ₁Q₂/d²
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F is the electrical force
k is a constant equal to 9×10⁹ 
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d is the distance between the two objects

Substituting the values:

F = (9×10⁹)(-22×10⁻⁹ C)(-22×10⁻⁹ C)/(0.10 m)²
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3 years ago
What is your angular position 75 seconds after the wheel starts turning, measured counterclockwise from the top? Express your an
AveGali [126]

Complete Question

A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the wheel starts turning, you are at the very top.

What is your angular position 75 seconds after the wheel starts turning, measured counterclockwise from the top? Express your answer as an angle between 0∘ and 360∘. Express your answer in degrees.

Answer:

\phi=123.75

Explanation:

From the question we are told that:

Height h=27m

Period T=32sec

Time t=75sec

Generally the equation for angular velocity is mathematically given by

\omega=\frac{2 \pi}{T}

\omega=\frac{2 \pi}{32}

\omega=0.196rad/s

Therefore

\theta=\omega t

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\phi=\theta-2(360)

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6 0
3 years ago
A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The
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Answer:

The tank is losing 4.976*10^{-4}  m^3/s

v_g = 19.81 \ m/s

Explanation:

According to the Bernoulli’s equation:

P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 +  \frac{1}{2}  \rho v_2^2 + \rho gh_2

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume  v_1 ≅ 0 ;

then v_2 can be determined as:\sqrt{[2g (h_1- h_2)]

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

v_2 = \sqrt{[2*9.81*(20 - 15)]

v_2 = \sqrt{[2*9.81*(5)]

v_2= 9.9 \ m/s  as it leaves the hole at the base.

radius r = d/2  = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = A_1v_2

J = πr²v_2    

J =\pi *(2*10^{-3})^{2}*9.9

J =1.244*10^{-4}  m^3/s

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : v_g = \sqrt{392.31}

v_g = 19.81 \ m/s

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Answer:

The distance between the camera and the rock is 836.6 cm

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cos α = adjacent/hypotenuse

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h = 198.0/cos(76.31)

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