The resulting change in momentum of the system will be +18.6 Ns. The momentum is conserved.
<h3>What is the law of conservation of momentum?</h3>
According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.
The given data in the problem is;
m is the mass =6.0 kg
t is the time interval=2 second
From Newton's second law;
From the graph;
The change in the momentum is;
Hence, the resulting change in momentum of the system will be +18.6 Ns.
To learn more about the law of conservation of momentum, refer;
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Answer:
c. The steady-state value of the current depends on the resistance of the resistor.
Explanation:
Since all the components are connected in series, when the switch is at first open, current will not flow round the circuit. As current needs to flow through from the positive terminal of the battery through the resistor, inductor, and switch to the negative terminal of the battery.
But the moment the switch is closed, at the initial time t = 0, the current flow through from the positive terminal of the battery through the resistor, inductor, and switch to the negative terminal of the battery. It then begins to increase at a rate that depends upon the value of the inductance of the inductor.
Answer:
B
Explanation:
Remark
If, when it reaches its maximum height where the PE is the greatest, that the maximum energy is when the ball is first kicked. That would mean its starting energy is 10 Joules.
Formula
KE = 1/2 m v^2
Givens
KE = 10 J
m = 0.2 kg
v = ?
Solution
10 = 1/2 * 0.2 * v^2
10 * 2 = 0.2 * v^2
20 = 0.2 * v^2
20/0.2 = v^2
100 = v^2
v = 10 m/s
Answer:
Explanation:
Velocity of Top most point of wheel is twice the Velocity of centre of mass of wheel
Thus angular velocity is given by
![K.E.=\frac{1}{2}m_{stone}V^2+2\left [ \frac{1}{2}m_{cyl}\left [ \frac{V}{2}\right ]^2\right ]+2\left [ \frac{1}{2}I\omega ^2\right ]](https://tex.z-dn.net/?f=K.E.%3D%5Cfrac%7B1%7D%7B2%7Dm_%7Bstone%7DV%5E2%2B2%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B2%7Dm_%7Bcyl%7D%5Cleft%20%5B%20%5Cfrac%7BV%7D%7B2%7D%5Cright%20%5D%5E2%5Cright%20%5D%2B2%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B2%7DI%5Comega%20%5E2%5Cright%20%5D)
![K.E.=\frac{V^2}{2}\left [ m_{stone}+\frac{m_{roller}}{2}\right ]+I\omega ^2](https://tex.z-dn.net/?f=K.E.%3D%5Cfrac%7BV%5E2%7D%7B2%7D%5Cleft%20%5B%20m_%7Bstone%7D%2B%5Cfrac%7Bm_%7Broller%7D%7D%7B2%7D%5Cright%20%5D%2BI%5Comega%20%5E2)
![K.E.=\frac{0.319^2}{2}\left [ 672+\frac{82}{2}\right ]+\frac{82\times 0.343^2\times 0.465^2}{2}](https://tex.z-dn.net/?f=K.E.%3D%5Cfrac%7B0.319%5E2%7D%7B2%7D%5Cleft%20%5B%20672%2B%5Cfrac%7B82%7D%7B2%7D%5Cright%20%5D%2B%5Cfrac%7B82%5Ctimes%200.343%5E2%5Ctimes%200.465%5E2%7D%7B2%7D)
