Answer:
a. 11 m/s at 76° with respect to the original direction of the lighter car.
Explanation:
In this exercise, since both cars make a right angle, let's assume that the lighter car only has a horizontal velocity component (vx) and that the heavier one only has a vertical velocity component (vy). The final velocities for both components for the system can be determined as:

Assume that the lighter car has a 1kg mass and that the heavier car has a 4 kg mass.

The magnitude of the final velocity of the wreck can be found as:
![v_{f}^{2}= v_{fx}^{2}+ v_{fy}^{2}\\v_{f}=\sqrt[]{2.6^{2} + 10.4^{2}} \\v_{f}= 10.72](https://tex.z-dn.net/?f=v_%7Bf%7D%5E%7B2%7D%3D%20v_%7Bfx%7D%5E%7B2%7D%2B%20v_%7Bfy%7D%5E%7B2%7D%5C%5Cv_%7Bf%7D%3D%5Csqrt%5B%5D%7B2.6%5E%7B2%7D%20%2B%2010.4%5E%7B2%7D%7D%20%5C%5Cv_%7Bf%7D%3D%2010.72)
The final velocity has an intensity of roughly 11 m/s
As for the angle, it can be determined in respect to the lighter car (x axis) as follows:

Therefore, the wreck has a velocity with an intensity of 11 m/s at 76° with respect to the original direction of the lighter car.
Answer:
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Answer:
16.26 cm in front of the mirror
Explanation:
Using,
1/f = 1/u+1/v....................... Equation 1
Where f = focal focal length of the concave mirror, u = object distance, v = image distance.
make v the subject of the equation
v = fu/(u-f)................... Equation 2
Note: The focal length of a concave mirror is positive
Using the real- is- positive convention
Given: f = 22/2 = 11 cm, u = 34 cm.
Substitute into equation 2
v = (34×11)/(34-11)
v = 374/23
v = 16.26 cm.
The image will be formed 16.26 cm in front of the mirror.