C. coil suspended by bearings.
<span>but im not 100% sure</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>
Answer:
Options A and D
Explanation:
In this question the student needs to collect these measurements in order to approximate the work done
A. The mass of the student
D. The final vertical height above the initial vertical position.
Workdone = mgh
m = mass
g = gravity = 9.8m/s²
h = vertical height between the initial and the final positions.
The vertical height has to be known as gravity only acts straight down.
Answer:
The momentum of bath cars is 40000 Ns which make the difficulty to stop each car in aspect of fprce is the same.
Explanation:
Momentum (P) =mass(m) × velocity (v)
For car A,
P = m × v = 1000 × 40 = 40000 Ns
For car B,
P = m × v = 4000 × 10 = 40000 Ns
Force (F) = Momentum change(ΔΡ)/ time taken(t)
F = ΔΡ/t
When stopping the car the momentum changes from 40000 Ns to 0
So momentum change in both cars is the same. So to stop the two cars in a given time (t) you need the same force, which means you will feel same difficulty.
