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lana66690 [7]
3 years ago
7

Describe how the ringing sound of a telephone travels from the phone to your ear

Physics
1 answer:
ehidna [41]3 years ago
6 0
The phone gives off sound waves, and the sound waves go into your ear to be interpreted by ur brain
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What do you call the maximum displacement of a wave from the equilibrium position?
weeeeeb [17]
Personally, I call it the wave's "amplitude".

What do YOU call it ?
7 0
3 years ago
Ondrea could drive a Jetson's flying car at a constant speed of 540.0 km/hr across oceans and space, approximately how long woul
Tcecarenko [31]

Answer:

The time taken in years is   x = 125 \  years

Explanation:

From the question we are told that

   The  speed is  v  = 540.00 \ km /hr  =  \frac{540 *1000}{3600} =  150\  m/s

    The distance from the sun to Pluto is  d =  5.9*10^{9} \  k m =  5.9*10^9 * 1000 =  5.9*10^{12} \  m

Generally the time  taken is mathematically represented as

     t =  \frac{d}{v}

=>   t = \frac{5.9*10^{11}}{150}

=>   t =  3.933*10^{9}

Converting to years

   1 year  \to  3.154*10^7 \  s

    x \  years  \to 3.933*10^{9}

=>  x = \frac{ 3.933*10^{9} * 1 }{ 3.154 *10^7}

=>    x = 125 \  years

7 0
3 years ago
Three equal point charges, each with charge 1.15 μCμC , are placed at the vertices of an equilateral triangle whose sides are of
julsineya [31]

Answer:

U=50.96J

Explanation:

The electrostatic potential energy for pair of charge is given by

U=1/4π∈₀×(q₁q₂/r)

Hence for a system of three charges the electrostatic potential energy can be found by adding up the potential energy for all possible pairs or charges.For three equal charges on the corners of an equilateral triangle,the electrostatic potential energy is given by:

U=1/4π∈₀×(q²/r)+1/4π∈₀×(q²/r)+1/4π∈₀×(q²/r)

U=3×1/4π∈₀×(q²/r)

Substitute given values

So

U=3\frac{1}{4\pi E_{o} }\frac{q^{2} }{r}\\  U=3\frac{1}{4\pi8.85*10^{-12} }\frac{(1.15*10^{-6}C )^{2} }{0.0007m}\\  U=50.96J

8 0
3 years ago
found the potential spring energy. not sure how to find the height of the block by knowing only the mass. found the weight of it
evablogger [386]

(h + .16) m g = 1/2 k x^2   total PE of block relative to where it stops

(h + .16) .82 * 9.8 = .5 * 120 * .16^2    PE released = PE of  spring

8.04 h + 1.29 = 1.536

h = (1.536 - 1.29) / 8.04 = .031 m = 3.1 cm

7 0
3 years ago
A constant voltage of 12.00 V has been observed over a certain time interval across a 1.20 H inductor. The current through the i
kiruha [24]

Answer:

The time is 0.5 sec.

Explanation:

Given that,

Voltage V= 12.00 V

Inductance L= 1.20 H

Current = 3.00 A

Increases rate = 8.00 A

We need to calculate change in current

\Delta A = 8.00-3.00= 5.00\ A

We need to calculate the time interval

Using formula of inductor

V=L\dfrac{\Delta A}{\Delta t}

\Delta t =\dfrac{L\Delta A}{V}

Where, \Delta A = change in current

V = voltage

L = inductance

Put the value into the formula

\Delta t=\dfrac{1.20\times5.00}{12.00}

\Delta t=0.5\ sec

Hence, The time is 0.5 sec.

5 0
3 years ago
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