I believe the answer is liquid.
Because the friction on the surface you are sliding on and the friction of the crate, go against each other. So once you start moving the friction will still be there but you now have the momentum to push past the starting point.
Answer:
F=(-4.8*10^22,0,0) N
Explanation:
<u>Given :</u>
We are given the magnitude of the momentum of the planet and let us call this momentum (p_now) and it is given by p_now = 2.60 × 10^29 kg·m/s. Also, we are given the force exerted on the planet F = 8.5 × 10^22 N. and the angle between the planet and the star is Ф = 138°
Solution :
We are asked to find the parallel component of the force F The momentum here is not constant, where the planet moving along a curving path with varying speed where the rate change in momentum and the force may be varying in magnitude and direction. We divide the force here into two parts: a parallel force F to the momentum and a perpendicular force F' to the momentum.
The parallel force exerted to the momentum will speed or reduce the velocity of the planet and does not change its moving line. Let us apply the direction cosines, we could obtain the parallel force as next
F=|F|cosФp (1)
Where the parallel force F is in the opposite direction of p as the angle between them is larger than 90°. Now we can plug our values for 0 and I F I into equation (1) to get the parallel force to the planet
F=|F|cosФp
=-4.8*10^22 N*p
<em>As this force is in one direction, we could get its vector as next </em>
F=(-4.8*10^22,0,0) N
F=(0,-4.8*10^22,0) N
F=(0,0-4.8*10^22) N
The cosine of 138°, the angle between F and p is, is a negative number, so F is opposite to p. The magnitude of the planet's momentum will decrease.
Answer:
A , power
Explanation:
Hope this is useful. Have a lovely rest of your day! God bless you.
