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2 years ago

A TV satellite dish is designed to receive radio waves of wavelength

Physics

1 answer:

Verizon [17]2 years ago

3 0

Answer:

4.7 GHz

Explanation:

Applying,

v = λf................. Equation 1

Where v = velocity of the radio wave, λ = wavelength, f = frequency

make f the subject of the equation

f = v/λ.............. Equation 2

Note: A radio wave is an electromagnetic wave, as such it moves with a velocity of 3.00 x 10⁸ m/s

From the question,

Given: λ = 0.0644 meters

Constant: v = <em>3.00 x 10⁸ m/s</em>

Substitute these values into equation 2

f = (3.00 x 10⁸)/0.0644

f = 4.66×10⁹ Hz

f = 4.7 GHz

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Two charges are located in the x–y plane. If q1 = -2.90 nC and is located at x = 0.00 m, y = 0.840 m and the second charge has m

Lunna [17]

Answer:

Epx= - 21.4N/C

Epy= 19.84N/C

Explanation:

Electric field theory

The electric field at a point P due to a point charge is calculated as follows:

E= k*q/r²

E= Electric field in N/C

q = charge in Newtons (N)

k= electric constant in N*m²/C²

r= distance from load q to point P in meters (m)

Equivalences

1nC= 10⁻⁹C

known data

q₁=-2.9nC=-2.9 *10⁻⁹C

q₂=5nC=5 *10⁻⁹C

r₁=0.840m

$r_{2} =\sqrt{1^{2} +0.8^{2} } =\sqrt{1.64}$

$sin\beta =\frac{0.8}{\sqrt{1.64} } =0.6246$

$cos\beta =\frac{1}{\sqrt{1.64} } =0.7808$

Calculation of the electric field at point P due to q1

Ep₁x=0

$Ep_{1y} =\frac{k*q_{1} }{r_{1}^{2} } =\frac{8.99*10^{9}*2.9*10^{-9} }{0.84^{2} } =36.95\frac{N}{C}$

Calculation of the electric field at point P due to q2

$Ep_{2x} =-\frac{k*q_{2} *cos\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.7808 }{(\sqrt{1.64})^{2} } =-21.4\frac{N}{C}$

$Ep_{2y} =-\frac{k*q_{2} *sin\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.6242 }{(\sqrt{1.64})^{2} } =-17.11\frac{N}{C}$

Calculation of the electric field at point P(0,0) due to q1 and q2

Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C

Epy= Ep₁y+ Ep₂y=36.95 N/C-17.11N =19.84N/C

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