Answer:
4v/3
Explanation:
Assume elastic collision by the law of momentum conservation:
![m_1v = m_1v_1 + m_2v_2](https://tex.z-dn.net/?f=m_1v%20%3D%20m_1v_1%20%2B%20m_2v_2)
where v is the original speed of car 1, v1 is the final speed of car 1 and v2 is final speed of car 2. m1 and m2 are masses of car 1 and car 2, respectively
Substitute ![m_2 = m_1/2 \& v_1 = v/3](https://tex.z-dn.net/?f=m_2%20%3D%20m_1%2F2%20%5C%26%20v_1%20%3D%20v%2F3)
![m_1v = \frac{m_1v}{3} + \frac{m_1v_2}{2}](https://tex.z-dn.net/?f=m_1v%20%3D%20%5Cfrac%7Bm_1v%7D%7B3%7D%20%2B%20%5Cfrac%7Bm_1v_2%7D%7B2%7D)
Divide both side by
, then multiply by 6 we have
![6v = 2v + 3v_2](https://tex.z-dn.net/?f=6v%20%3D%202v%20%2B%203v_2)
![3v_2 = 4v](https://tex.z-dn.net/?f=3v_2%20%3D%204v)
![v_2 = \frac{4v}{3}](https://tex.z-dn.net/?f=v_2%20%3D%20%5Cfrac%7B4v%7D%7B3%7D)
So the final speed of the second car is 4/3 of the first car original speed
The answer is B. magnesium I am pretty sure
D. The red car is moving faster than the blue car
Answer:
540C.
Explanation:
A capacitor of capacitance C when charged to a voltage of V will have a charge Q given as follows;
Q = CV ----------(i)
From the question, the initial charge on the capacitor is the charge on it before it was connected to the resistor. In other words, the initial charge on the capacitor will have a maximum value which can be calculated using equation (i) above.
Where;
C = 6F
V = 90V
Substitute these values into equation (i) as follows;
Q = 6 x 90
Q = 540 C
Therefore, the initial charge on the capacitor is 540C.