Answer:
When two forces acting on an object are equal in size but act in opposite directions, we say that they are balanced forces.
Explanation:
Answer:
![v = 9.88 \times 10^7 m/s](https://tex.z-dn.net/?f=v%20%3D%209.88%20%5Ctimes%2010%5E7%20m%2Fs)
Explanation:
As we know that electrostatic potential energy of the electron is converting here into kinetic energy
so we will have
![qV = \frac{1}{2}mv^2](https://tex.z-dn.net/?f=qV%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
so here we know that
![q = 1.6 \times 10^{-19} C](https://tex.z-dn.net/?f=q%20%3D%201.6%20%5Ctimes%2010%5E%7B-19%7D%20C)
![V = 27800 Volts](https://tex.z-dn.net/?f=V%20%3D%2027800%20Volts)
![m = 9.1 \times 10^{-31} kg](https://tex.z-dn.net/?f=m%20%3D%209.1%20%5Ctimes%2010%5E%7B-31%7D%20kg)
so from above equation we have
![v = \sqrt{\frac{2qV}{m}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7B2qV%7D%7Bm%7D%7D)
![v = \sqrt{\frac{2(1.6 \times 10^{-19})(27800)}{9.11 \times 10^{-31}}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7B2%281.6%20%5Ctimes%2010%5E%7B-19%7D%29%2827800%29%7D%7B9.11%20%5Ctimes%2010%5E%7B-31%7D%7D%7D)
![v = 9.88 \times 10^7 m/s](https://tex.z-dn.net/?f=v%20%3D%209.88%20%5Ctimes%2010%5E7%20m%2Fs)
Answer:
a) ![v_{3} =8.43 m/s](https://tex.z-dn.net/?f=v_%7B3%7D%20%3D8.43%20m%2Fs)
b) ![v_{2}=2.15m/s](https://tex.z-dn.net/?f=v_%7B2%7D%3D2.15m%2Fs)
c) ΔK=![-28.18x10^4J](https://tex.z-dn.net/?f=-28.18x10%5E4J)
d)ΔK=![-10.33x10^4J](https://tex.z-dn.net/?f=-10.33x10%5E4J)
Explanation:
From the exercise we know that there is a collision of a sports car and a truck.
So, the sport car is going to be our object number 1 and the truck object number 2.
![m_{1}=1050kg\\v_{1}=-13m/s\\m_{2}=6320kg\\v_{2}=12m/s](https://tex.z-dn.net/?f=m_%7B1%7D%3D1050kg%5C%5Cv_%7B1%7D%3D-13m%2Fs%5C%5Cm_%7B2%7D%3D6320kg%5C%5Cv_%7B2%7D%3D12m%2Fs)
Since the two vehicles remain locked together after the collision the final mass is:
![m_{3}=7370kg](https://tex.z-dn.net/?f=m_%7B3%7D%3D7370kg)
a) To find the velocity of the two vehicles just after the collision we must use linear's momentum principle
![p_{1}=p_{2}](https://tex.z-dn.net/?f=p_%7B1%7D%3Dp_%7B2%7D)
![m_{1}v_{1}+ m_{2}v_{2}=m_{3}v_{3}](https://tex.z-dn.net/?f=m_%7B1%7Dv_%7B1%7D%2B%20m_%7B2%7Dv_%7B2%7D%3Dm_%7B3%7Dv_%7B3%7D)
![v_{3}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{3}}=\frac{(1050kg)(-13m/s)+(6320kg)(12m/s)}{7370kg}](https://tex.z-dn.net/?f=v_%7B3%7D%3D%5Cfrac%7Bm_%7B1%7Dv_%7B1%7D%2Bm_%7B2%7Dv_%7B2%7D%7D%7Bm_%7B3%7D%7D%3D%5Cfrac%7B%281050kg%29%28-13m%2Fs%29%2B%286320kg%29%2812m%2Fs%29%7D%7B7370kg%7D)
![v_{3}=8.43m/s](https://tex.z-dn.net/?f=v_%7B3%7D%3D8.43m%2Fs)
b) To find the speed the truck should have had so both vehicles stopped in the collision we need to use the same principle used before
![m_{1}v_{1}+ m_{2}v_{2}=0](https://tex.z-dn.net/?f=m_%7B1%7Dv_%7B1%7D%2B%20m_%7B2%7Dv_%7B2%7D%3D0)
![v_{2}=\frac{-m_{1}v_{1}}{m_{2} }=\frac{-(1050kg)(-13m/s)}{(6320kg)}=2.15m/s](https://tex.z-dn.net/?f=v_%7B2%7D%3D%5Cfrac%7B-m_%7B1%7Dv_%7B1%7D%7D%7Bm_%7B2%7D%20%7D%3D%5Cfrac%7B-%281050kg%29%28-13m%2Fs%29%7D%7B%286320kg%29%7D%3D2.15m%2Fs)
c) To find the change in kinetic energy we need to do the following steps:
ΔK=![k_{2}-k_{1}=\frac{1}{2}m_{3}v_{3}^{2}-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2} )](https://tex.z-dn.net/?f=k_%7B2%7D-k_%7B1%7D%3D%5Cfrac%7B1%7D%7B2%7Dm_%7B3%7Dv_%7B3%7D%5E%7B2%7D-%28%5Cfrac%7B1%7D%7B2%7Dm_%7B1%7Dv_%7B1%7D%5E%7B2%7D%2B%5Cfrac%7B1%7D%7B2%7Dm_%7B2%7Dv_%7B2%7D%5E%7B2%7D%20%29)
ΔK=![\frac{1}{2}(7370)(8.43)^{2}-(\frac{1}{2}(1050)(-13)^{2}+\frac{1}{2}(6320)(12)^{2} )=-28.18x10^{4}J](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%287370%29%288.43%29%5E%7B2%7D-%28%5Cfrac%7B1%7D%7B2%7D%281050%29%28-13%29%5E%7B2%7D%2B%5Cfrac%7B1%7D%7B2%7D%286320%29%2812%29%5E%7B2%7D%20%29%3D-28.18x10%5E%7B4%7DJ)
d) The change in kinetic energy where the two vehicles stopped in the collision is:
ΔK=![k_{2}-k_{1}=0-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2} )](https://tex.z-dn.net/?f=k_%7B2%7D-k_%7B1%7D%3D0-%28%5Cfrac%7B1%7D%7B2%7Dm_%7B1%7Dv_%7B1%7D%5E%7B2%7D%2B%5Cfrac%7B1%7D%7B2%7Dm_%7B2%7Dv_%7B2%7D%5E%7B2%7D%20%29)
ΔK=![-(\frac{1}{2}(1050)(-13)^{2}+\frac{1}{2}(6320)(2.15)^{2} )=-10.33x10^4J](https://tex.z-dn.net/?f=-%28%5Cfrac%7B1%7D%7B2%7D%281050%29%28-13%29%5E%7B2%7D%2B%5Cfrac%7B1%7D%7B2%7D%286320%29%282.15%29%5E%7B2%7D%20%29%3D-10.33x10%5E4J)
The answer will be Amplitude.