Answer:a) 34.5 N; b) 24.5 N; c) 10 N; d) 1J
Explanation: In order to solve this problem we have to used the second Newton law given by:
∑F= m*a
F-f=m*a where f is the friction force (uk*Normal), from this we have
F= m*a+f=5 Kg*2 m/s^2+0.5*5Kg*9.8 m/s^2= 34.5 N
then f=uk*N=0.5*5Kg*9.8 m/s^2= 24.5N
the net Force = (34.5-24.5)N= 10 N
Finally the work done by the net force is equal to kinetic energy change so
W=∫Force net*dr= 10 N* 0.1 m= 1J
First, foremost, and most critically, you must look at the graph, and critically
examine its behavior from just before until just after the 5-seconds point.
Without that ability ... since the graph is nowhere to be found ... I am hardly
in a position to assist you in the process.
The solid, liquid and gas phases of water would have the same structure of the molecules since they are same substance. The only difference would be the distances of the molecules in the container. For a ice, the molecules are close to each other where the molecules vibrate only in place. For liquid, the molecules are freely moving and are at some distance with each other but not that far away with each other. Steam, on the other hand, would have molecules that are very far from each other and are freely moving in the whole container. As the container is heated, the size of the molecules would not change. It is only the volume that has changed. Also, the mass is the same since there is no outflow of the substances.
![m_1=2 \\ m_2=3 \\ v_1=4 \\ v_2=1 \\ v\text{ =speed after collision (to be determined)}.](https://tex.z-dn.net/?f=m_1%3D2%20%5C%5C%20m_2%3D3%20%5C%5C%20v_1%3D4%20%5C%5C%20v_2%3D1%20%5C%5C%20v%5Ctext%7B%20%3Dspeed%20after%20collision%20%28to%20be%20determined%29%7D.)
The momentul of the system preserves:
Ok, we found the speed after the collision.
Now, because the impact is plastic, it produces heat, sound energy and who knows what other forms of energy. We denote all this wasted energy with
![E](https://tex.z-dn.net/?f=E)
.
Now, we write the energy conservation law:
![\dfrac{m_1v_1^2}{2}+\dfrac{m_2v^2_2}{2}=\dfrac{(m_1+m_2)v^2}{2}+E](https://tex.z-dn.net/?f=%5Cdfrac%7Bm_1v_1%5E2%7D%7B2%7D%2B%5Cdfrac%7Bm_2v%5E2_2%7D%7B2%7D%3D%5Cdfrac%7B%28m_1%2Bm_2%29v%5E2%7D%7B2%7D%2BE)
From the above equation, you find
![E](https://tex.z-dn.net/?f=E)
, and then conclude that the sound energy can certainly not be greater than this.