5.51 × 10 power 12 newton is answer
We assume that horn releases sound of constant frequency. In order for observer to observe different frequency either horn or observer or both must move.
This happens due to Doppler effect. It states that when position of source of sound and observer relative to each other changes, the observed frequency also changes. If the source emits sound of constant frequency than observed frequency will be either higher or lower than original.
When distance between source and observer increases the observed frequency will be lower. This is because same number of sound waves must cover greater distance so they have greater wavelength.
When distance between source and observer decreases the observed frequency will be higher. This is because same number of sound waves must cover smaller distance so they have smaller wavelength.
Wavelength and frequency are inversely proportional meaning when one increases the other drecreases.
From this explanation we can find answer for our question. <span>If we wanted the pitch of a horn to drop relative to an observer we need to move horn away from an observer.</span>
<span>The true brightness of an object
is called its luminosity. It is the total amount of energy emitted by bright or
meteorological objects over a period of time. It has the SI unit of joules per
second or watts. So the answer is letter A. Intensity is the measure of how
strong the substance or object is when it projects something. Magnitude is a
measure of how great is the size the object produces. Viscosity is the measure
of flow of a substance.</span>
Find the electric flux and the disp at t=0.50ns
<span>Given: </span>
<span>Resistor R = 160 Ω </span>
<span>Voltage ε = 22.0 V </span>
<span>Capacitor C = 3.10 pF = 3.10 * 10^-12 F </span>
<span>time t = 0.5 ns = 0.5 * 10^-9 s </span>
<span>ε0 = 8.85 * 10^-12 </span>
<span>Solution: </span>
<span>ELECTRIC FLUX: </span>
<span>Φ = Q/ε0 </span>
<span>we have ε0, we need to find Q the charge </span>
<span>STEP 1: FIND Q </span>
<span>Q = C ε ( 1 - e^(-t/RC) ) </span>
<span>Q = { 3.10 * 10^-12 } { 22.0 } { 1 - e^(- 0.5 * 10^-9 / 160 *3.10 * 10^-12 ) } </span>
<span>Q = { 3.10 * 10^-12 } { 22.0 } { 1 - 0.365 } </span>
<span>Q = { 3.10 * 10^-12 } { 22.0 } { 0.635 } </span>
<span>Q = 43.31 * 10^-12 C </span>
<span>STEP 2: WE HAVE Q AND ε0 > >>> SOLVE FOR ELECTRIC FLUX >>> </span>
<span>Φ = Q/ε0 </span>
<span>Φ = { 43.31 * 10^-12 C } / { ε0 = 8.85 * 10^-12 } </span>
<span>Φ = 4.8937 = 4.9 V.m </span>
<span>DISPLACEMENT CURRENT </span>
<span>we use the following equation: </span>
<span>I = { ε / R } { e^(-t/RC) } </span>
<span>I = { 22 / 160 } { e^(- 0.5 * 10^-9 / 160 *3.10 * 10^-12 ) } </span>
<span>I = { 0.1375 } { 0.365 } </span>
<span>I = 0.0502 A = 0.05 A </span>
