To resolve point A and B we need the concepts related to conservation of momentum (By collision) and Kinetic Energy. Conservation of momentum is given by the equation,
![m_1\vec{v_1}+m_2\vec{v_2} = (m_1+m_2)\vec{v}](https://tex.z-dn.net/?f=m_1%5Cvec%7Bv_1%7D%2Bm_2%5Cvec%7Bv_2%7D%20%3D%20%28m_1%2Bm_2%29%5Cvec%7Bv%7D)
Our values in the statment are:
![m_1 = 1200kg](https://tex.z-dn.net/?f=m_1%20%3D%201200kg)
![v_1 = 6m/s](https://tex.z-dn.net/?f=v_1%20%3D%206m%2Fs)
![m_2 = 900kg](https://tex.z-dn.net/?f=m_2%20%3D%20900kg)
![v_2 = 24m/s](https://tex.z-dn.net/?f=v_2%20%3D%2024m%2Fs)
Part A) As it is in an icy intersection, there is two different components (x,y) then,
![1200(-6\hat{j})+900(-24\hat{i}) = (1200+900)\vec{v}](https://tex.z-dn.net/?f=1200%28-6%5Chat%7Bj%7D%29%2B900%28-24%5Chat%7Bi%7D%29%20%3D%20%281200%2B900%29%5Cvec%7Bv%7D)
![2100\vec{v} = -21600\hat{i}-7200\hat{j}](https://tex.z-dn.net/?f=2100%5Cvec%7Bv%7D%20%3D%20-21600%5Chat%7Bi%7D-7200%5Chat%7Bj%7D)
![\vec{v} = -72/7\hat{i}-24/7\hat{j}](https://tex.z-dn.net/?f=%5Cvec%7Bv%7D%20%3D%20-72%2F7%5Chat%7Bi%7D-24%2F7%5Chat%7Bj%7D)
Then the magnitude is,
![|\vec{v}| = 9.6525m/s](https://tex.z-dn.net/?f=%7C%5Cvec%7Bv%7D%7C%20%3D%209.6525m%2Fs)
Part B) To obtain the Kinetic Energy Loss we need to use its equation, which is given by,
![KE_i = \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2](https://tex.z-dn.net/?f=KE_i%20%3D%20%5Cfrac%7B1%7D%7B2%7Dm_1v_1%5E2%2B%5Cfrac%7B1%7D%7B2%7Dm_2v_2%5E2)
![KE_i = \frac{1}{2}(1200)(6)^2+\frac{1}{2}(900)(24)^2](https://tex.z-dn.net/?f=KE_i%20%3D%20%5Cfrac%7B1%7D%7B2%7D%281200%29%286%29%5E2%2B%5Cfrac%7B1%7D%7B2%7D%28900%29%2824%29%5E2)
![KE_i = 280.8kJ](https://tex.z-dn.net/?f=KE_i%20%3D%20280.8kJ)
The final energy is given by,
![KE_f = \frac{1}{2}(m_1+m_2)v_f^2](https://tex.z-dn.net/?f=KE_f%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28m_1%2Bm_2%29v_f%5E2)
![KE_f = \frac{1}{2} (1200+900)(9.65)](https://tex.z-dn.net/?f=KE_f%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%281200%2B900%29%289.65%29)
![KE_f =97778.625J](https://tex.z-dn.net/?f=KE_f%20%3D97778.625J)
Then the change in Kinetic Energy is
![\Delta KE = KE_f-KE_i = 97.778kJ- 280.8kJ](https://tex.z-dn.net/?f=%5CDelta%20KE%20%3D%20KE_f-KE_i%20%3D%2097.778kJ-%20280.8kJ)
![\Delta KE = -183.02kJ](https://tex.z-dn.net/?f=%5CDelta%20KE%20%3D%20-183.02kJ)
<em>There was a loss of KE of 183.02kJ</em>
Answer:
"Offgassing"
Explanation:
According to my research on Kinesiology, I can say that based on the information provided within the question the process being described is known as "Offgassing". In other words this process is defined as when something gives off or releases a chemical, especially a harmful one, in the form of a gas into the air..
I hope this answered your question. If you have any more questions feel free to ask away at Brainly.
Answer:
The officer's unit detects this 135-mile-per-hour speed and should subtract the patrol car's 70-mile -per-hour ground speed to get your true speed of 65 miles per hour. Instead, the officer's ground-speed beam fixes on the truck ahead and measures a false 50-mile-per-hour ground speed.
Explanation:
A speedometer or speed meter is a gauge that measures and displays the instantaneous speed of a vehicle. Now universally fitted to motor vehicles, they started to be available as options in the early 20th century, and as standard equipment from about 1910 onwards.
Where's the diagram for question 1?