Answer:
Relative dating methods are unable to determine the absolute age of an object or event, but can determine the impossibility of a particular event happening before or after another event of which the absolute date is well known.
So, Maria is using relative dating method
It's the only answer
An example of using muscular strength would be doing things like lifting weights and things like that.
Answer:
<em>Amplitude= 8 m</em>
Explanation:
<u>The Amplitude of a Wave</u>
Sinusoidal Function refers to a mathematical curve with a smooth and periodic oscillation. Its name comes from the sine function and is characterized by the amplitude or the maximum displacement or distance moved by a point on a vibrating body measured from its equilibrium position.
To calculate the amplitude from a graph, we measure the maximum point and the minimum point the wave reaches. Then we subtract both values and divide the result by 2.
The shown wave in the figure has a maximum value of 8 m and a minimum value of -8 m. The distance from the maximum to the minimum is 8-(-8)= 16 m, thus the amplitude is 16/2= 8m.
Amplitude= 8 m
Answer:
a.)
Explanation:
a.) is the only possible answer because it is in m/s^2 which are the units for acceleration, and you already know the magnitude of acceleration will be 5 given the other options.
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>
