The correct answer is (c.) resists a change in motion of an object. Inertia basically resists any physical changes in an object in terms of its state. As stated in the Newton's first law of motion or also known as law of inertia, an object that is at rest will stay at rest and an object that is in motion will stay in motion. In short, an object will keep doing what it is already doing UNLESS net force is acted upon it,
Answer:
Leaves are the primary source of photosynthesis. Leaves are also have a vital function in plant processes like transpiration and guttation. I hope it helps:)
Answer:The gravitational force on the ball due to earth is exactly the same as the gravitational force on earth due to the ball
Explanation:
The gravitational force exist in pair. It form action-reaction pair.
B) equal in magnitude but opposite direction
Answer:
![V_f = 13.8863 \angle 60.89\°](https://tex.z-dn.net/?f=V_f%20%3D%2013.8863%20%5Cangle%2060.89%5C%C2%B0)
Explanation:
Our values are,
![m_1 = 950Kg\\v_1 = 16m/s \\m_2 =1300Kg\\v_2 = 21m/s](https://tex.z-dn.net/?f=m_1%20%3D%20950Kg%5C%5Cv_1%20%3D%2016m%2Fs%20%5C%5Cm_2%20%3D1300Kg%5C%5Cv_2%20%3D%2021m%2Fs)
We have all the values to apply the law of linear momentum, however, it is necessary to define the two lines in which the study will be carried out. Being an intersection the vehicle of mass m_1 approaches through the X axis, while the vehicle of mass m_2 approaches by the y axis. In the collision equation on the X axis, we despise the velocity of object 2, since it does not come in this direction.
![m_1v_1=(m_1+m_2)v_fcos\theta](https://tex.z-dn.net/?f=m_1v_1%3D%28m_1%2Bm_2%29v_fcos%5Ctheta)
For the particular case on the Y axis, we do the same with the speed of object 1.
![m_2v_2=(m_1+m_2)v_fsin\theta](https://tex.z-dn.net/?f=m_2v_2%3D%28m_1%2Bm_2%29v_fsin%5Ctheta)
By taking a final velocity as a component, we can obtain the angle between the two by relating the equations through the tangent
![Tan\theta = \frac{m_2v_2}{m_1v_1}\\Tan\theta = \frac{1300*21}{950*16}\\\theta = tan^{-1}(1.7960)\\\theta = 60.89\°](https://tex.z-dn.net/?f=Tan%5Ctheta%20%3D%20%5Cfrac%7Bm_2v_2%7D%7Bm_1v_1%7D%5C%5CTan%5Ctheta%20%3D%20%5Cfrac%7B1300%2A21%7D%7B950%2A16%7D%5C%5C%5Ctheta%20%3D%20tan%5E%7B-1%7D%281.7960%29%5C%5C%5Ctheta%20%3D%2060.89%5C%C2%B0)
Replacing in any of the two functions, given above, we will find the final speed after the collision,
![(950)(16)=(950+1300)V_fcos(60.89)](https://tex.z-dn.net/?f=%28950%29%2816%29%3D%28950%2B1300%29V_fcos%2860.89%29)
![V_f= \frac{(950)(16)}{(950+1300)cos(60.89)}](https://tex.z-dn.net/?f=V_f%3D%20%5Cfrac%7B%28950%29%2816%29%7D%7B%28950%2B1300%29cos%2860.89%29%7D)
![V_f = 13.8863 \angle 60.89\°](https://tex.z-dn.net/?f=V_f%20%3D%2013.8863%20%5Cangle%2060.89%5C%C2%B0)