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3 years ago

A 15,000-watt radio station transmits at 1700 kHz.

Physics

2 answers:

viva [34]3 years ago

6 0

the answer is a

i had this on a quiz

notsponge [240]3 years ago

3 0

$E_{\gamma}=hf= (6.62\times10^-^3^4 Js)(1700\times10^3Hz)=1.12\times10^-^2^7J/photon$
Answer A

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If a ball is tossed straight up into the air, at what position is its potential energy the greatest? Question 2 options: a) when

yarga [219]

Q1: At the highest point

3 0

4 years ago

A microphone is located on the line connecting two speakers that are 0.513 m apart and oscillating in phase. The microphone is 1

pantera1 [17]

Answer:

frequency 1 = 334.30 Hz

frequency 2 = 1002.92 Hz

Explanation:

Given data

speaker distance y = 0.513 m

microphone distance D = 1.80 m

to find out

lowest two frequencies

solution

we know velocity of sound is 343 m/s

so we consider point x

so at 1st speaker distance from x = D + (y/2)

1st speaker distance from x = 1.80 + (0.513/2) = 2.0565 m .....1

and

at 2nd speaker distance from x = D - (y/2)

2nd speaker distance from x = 1.80 - (0.513/2) = 1.5435 m .........2

so destructive interference from 1 and 2 we know

1st - 2nd = ( m + 0.5 ) wavelength

2.0565 m - 1.5435 m = ( 0+ 0.5) wavelength

wavelength = 1.026 m

so here 1st min frequency will be

frequency 1 = velocity of sound / wavelength

frequency 1 = 343 / 1.026 =334.30 Hz

and

2nd min frequency will be

frequency 2 =

2.0565 m - 1.5435 m = ( 1 + 0.5) wavelength

wavelength = 0.342 m

frequency 2 = velocity of sound / wavelength

frequency 2 = 343 / 0.342 = 1002.92 Hz

7 0

3 years ago

For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro

viktelen [127]

Answer:

$V_d$ = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, $A = \frac{\pi d^2}{4}$ = $A = \frac{\pi 0.625^2}{4}$

Now,

The current density, J is given as

$J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}$ = 2444619.925 A/mm²

now, the electron density, $n = \frac{\rho}{M}N_A$

where,

$N_A$ =Avogadro's Number

$n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3$

Now,

the drift velocity, $V_d$

$V_d=\frac{J}{ne}$

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

$V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e}$ = 1.75 × 10⁻⁴ m/s

4 0

3 years ago

Read 2 more answers

A 1500 kg car, initially traveling at 22.0 m/s, hits its brakes and skids to a stop. Determine the work done by friction.

Sedbober [7]

Answer:

The work done by the car is 363 kJ

Explanation:

Work : Work is said to be done when a Force moves an object through a certain distance. Work and Energy are interchangeable because they have the same unit. The unit of work is Joules (J).

Mathematically work done can be expressed as,

E = W = 1/2mv²

W = 1/2mv²................................ Equation 1

Where E = Energy, W = work done, m = mass of the car, v = velocity of the car

Given: m=1500 kg, v=22 m/s

Substituting these values into equation 1

W = 1/2(1500)(22)²

W = 750 × 484

W = 363000 J

W = 363 kJ

Thus the work done by the car is 363 kJ

8 0

3 years ago

Briefly explain how a resonance tube works

Lera25 [3.4K]

Answer:

As the tines of the tuning fork vibrate at their own natural frequency, they created sound waves that impinge upon the opening of the resonance tube. These impinging sound waves produced by the tuning fork force air inside of the resonance tube to vibrate at the same frequency.

4 0

2 years ago

Read 2 more answers