The answer is apparent weight is zero.
You are still accelerating downwards at 9.8m/s^2 (if you are on Earth).
You still are being affected by the Earth's gravity.
Not all because of the previous two statements.
Not none because apparent weight is zero as you are falling.
The way he got the wrong guy to get back into the car he was just saying that I didn’t need it for me so he said no I’m gonna do tell you I have a little girl that was going on a little bit of a girl and I
Winter
It is winter because the Northern Hemisphere is not facing the sun.
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.