Answer:
a)
, with a magnitude of 
b)
, with a magnitude of 
c)
, with a magnitude of 
Explanation:
We have:

We can calculate each component of the acceleration using its definition 

The rate of change of momentum of the ball is 
So for each coordinate:

And these are equal to the components of the net force since F=ma.
If magnitudes is what is asked:

<em>(N and </em>
<em> are the same unit).</em>
The centripetal force exerted on the automobile while rounding the curve is 
<u>Explanation:</u>
given that

Objects moving around a circular track will experience centripetal force towards the center of the circular track.

The answer is C. Watson and Crick developed the Double Helix model seen in the diagram.
Answer:
most evaporation and precipitation in the water cycle occus over the ocean
Answer:
60°
Explanation:
It will be equilateral triangle, you can prove with the cosine theorem.
1)P^2=P^2+P^2 - 2×P×P× cos A
2) cos A= P^2 / 2× P^2 = 1/2
cos A = 1/2
A = 60°
If you wrote this question right,it will be solved like this :)