Explanation:
When,the vehicle has uniform velocity, it's acceleration becomes zero
Answer:



Explanation:
= Uncertainty in position = 1.9 m
= Uncertainty in momentum
h = Planck's constant = 
m = Mass of object
From Heisenberg's uncertainty principle we know

The minimum uncertainty in the momentum of the object is 
Golf ball minimum uncertainty in the momentum of the object

Uncertainty in velocity is given by

The minimum uncertainty in the object's velocity is 
Electron


The minimum uncertainty in the object's velocity is
.
Answer:
Hindi ko alam pasensya ka ha godbless
Answer:
Fr = 26.83 [N]
Explanation:
To solve this problem we must use the Pythagorean theorem, since the forces are vector quantities, that is, they have magnitude and density. Therefore the Pythagorean theorem is suitable for the solution of this problem.
![F_{r}=\sqrt{(12)^{2}+(24)^{2} } \\F_{r}=26.83[N]](https://tex.z-dn.net/?f=F_%7Br%7D%3D%5Csqrt%7B%2812%29%5E%7B2%7D%2B%2824%29%5E%7B2%7D%20%20%7D%20%5C%5CF_%7Br%7D%3D26.83%5BN%5D)