Answer:
32500 kg m/s east, 42500 kg m/s west. Second car has larger momentum
Explanation:
The momentum of an object is given by
p = mv
where
m is the mass
v is the velocity
For the first car, m = 500 kg and v = 65 m/s east, so the momentum is
east
For the second car, m = 500 kg and v = 85 m/s west, so the momentum is
west
By comparing the two momentum, we see that the second car has larger momentum.
precambrian is the correct answer
Answer:
![v_f=9,07~m/s](https://tex.z-dn.net/?f=v_f%3D9%2C07~m%2Fs)
Explanation:
<u>Constant Acceleration Motion</u>
It's a type of motion in which the velocity of an object changes uniformly in time.
Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:
![v_f=v_o+at](https://tex.z-dn.net/?f=v_f%3Dv_o%2Bat)
The car initially travels at vo=7.35 m/s and accelerates at a rate of
during t=2.09 s.
The final velocity is:
![v_f=7.35+0.824*2.09](https://tex.z-dn.net/?f=v_f%3D7.35%2B0.824%2A2.09)
![\mathbf{v_f=9,07~m/s}](https://tex.z-dn.net/?f=%5Cmathbf%7Bv_f%3D9%2C07~m%2Fs%7D)
Complete Question
The complete question(reference (chegg)) is shown on the first uploaded image
Answer:
The magnitude of the resultant force is ![F = 199.64 \ N](https://tex.z-dn.net/?f=F%20%20%3D%20%20199.64%20%5C%20N)
The direction of the resultant force is
from the horizontal plane
Explanation:
Generally when resolving force, if the force (F )is moving toward the angle then the resolve force will be
while if the force is moving away from the angle then the resolved force is ![Fsin (\theta )](https://tex.z-dn.net/?f=Fsin%20%28%5Ctheta%20%29)
Now from the diagram let resolve the forces to their horizontal component
So
![\sum F_x = 150 cos(30) + 100cos(15) -80sin (20)](https://tex.z-dn.net/?f=%5Csum%20F_x%20%20%3D%20%20150%20cos%2830%29%20%2B%20100cos%2815%29%20-80sin%20%2820%29)
![\sum F_x = 199.128 \ N](https://tex.z-dn.net/?f=%5Csum%20F_x%20%20%3D%20%20199.128%20%5C%20N)
Now resolving these force into their vertical component can be mathematically evaluated as
![\sum F_{y} = 150 sin(30) - 100sin(15) -110 +80 cos(20)](https://tex.z-dn.net/?f=%5Csum%20%20F_%7By%7D%20%20%3D%20%20150%20sin%2830%29%20-%20100sin%2815%29%20-110%20%2B80%20cos%2820%29)
![\sum F_{y} = 14.30](https://tex.z-dn.net/?f=%5Csum%20%20F_%7By%7D%20%20%3D%20%2014.30)
Now the resultant force is mathematically evaluated as
![F = \sqrt{F_x^2 + F_y^2}](https://tex.z-dn.net/?f=F%20%20%3D%20%20%5Csqrt%7BF_x%5E2%20%2B%20F_y%5E2%7D)
substituting values
![F = \sqrt{199.128^2 + 14.3^2}](https://tex.z-dn.net/?f=F%20%20%3D%20%20%5Csqrt%7B199.128%5E2%20%2B%2014.3%5E2%7D)
![F = 199.64 \ N](https://tex.z-dn.net/?f=F%20%20%3D%20%20199.64%20%5C%20N)
The direction of the resultant force is evaluated as
![\theta = tan^{-1}[\frac{F_y}{F_x} ]](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%20tan%5E%7B-1%7D%5B%5Cfrac%7BF_y%7D%7BF_x%7D%20%5D)
substituting values
![\theta = tan^{-1}[\frac{ 14.3}{199.128} ]](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%20tan%5E%7B-1%7D%5B%5Cfrac%7B%2014.3%7D%7B199.128%7D%20%5D)
from the horizontal plane