In this question a lot of information's are provided. Among the information's provided one information and that is the time of 4 seconds is not required for calculating the answer. Only the other information's are required.
Mass of the block that is sliding = 5.00 kg
Distance for which the block slides = 10 meters/second
Then we already know that
Momentum = Mass * Distance travelled
= (5 * 10) Kg m/s
= 50 kg m/s
So the magnitude of the blocks momentum is 50 kg m/s. The correct option among all the given options is option "b".
Consider two variables said to be "inversely proportional" to each other. If all other variables are held constant, the magnitude or absolute value of one inversely proportional variable decreases if the other variable increases, while their product (the constant of proportionality k) is always the same.
Answer:
The maximum speed that the truck can have and still be stopped by the 100m road is the speed that it can go and be stopped at exactly 100m. Since there is no friction, this problem is similar to a projectile problem. You can think of the problem as being a ball tossed into the air except here you know the highest point and you are looking for the initial velocity needed to reach that point. Also, in this problem, because there is an incline, the value of the acceleration due to gravity is not simply g; it is the component of gravity acting parallel to the incline. Since we are working parallel to the plane, also keep in mind that the highest point is given in the problem as 100m. Solving for the initial velocity needed to have the truck stop after 100m, you should find that the maximum velocity the truck can have and be stopped by the road is 18.5 m/s.
Explanation:
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>