Answer:
A ball is thrown straight up with a speed of 30
m/s. What is the maximum height reached by
the ball?
I think the question meant to say net force on the box. Since force is a vector, the direction matters. 20N left is negated completely by the 50N right, which means the net force is 50N-20N to the right, 30N.
Answer:
≅50°
Explanation:
We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:
Δx=V₀t+at²/2
And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:
Δx=(V₀cosθ)t+at²/2
Now luckily we are given everything we need to solve (or you found the info before posting here):
- Δx=760 m
- V₀=87 m/s
- t=13.6 s
- a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!
With that we can plug the values in to get:
![760=(87)(cos\theta )(13.6)+\frac{(0)(13.6^{2}) }{2}](https://tex.z-dn.net/?f=760%3D%2887%29%28cos%5Ctheta%20%29%2813.6%29%2B%5Cfrac%7B%280%29%2813.6%5E%7B2%7D%29%20%7D%7B2%7D)
![760=(1183.2)(cos\theta)](https://tex.z-dn.net/?f=760%3D%281183.2%29%28cos%5Ctheta%29)
![cos\theta=\frac{760}{1183.2}](https://tex.z-dn.net/?f=cos%5Ctheta%3D%5Cfrac%7B760%7D%7B1183.2%7D)
![\theta=cos^{-1}(\frac{760}{1183.2})\approx50^{o}](https://tex.z-dn.net/?f=%5Ctheta%3Dcos%5E%7B-1%7D%28%5Cfrac%7B760%7D%7B1183.2%7D%29%5Capprox50%5E%7Bo%7D)
Answer:
160N
Explanation:
Moments must be conserved - so.
![20 * 0.4 = F * 0.05](https://tex.z-dn.net/?f=20%20%2A%200.4%20%3D%20F%20%2A%200.05)
![F=\frac{20*0.4}{0.05} = 160](https://tex.z-dn.net/?f=F%3D%5Cfrac%7B20%2A0.4%7D%7B0.05%7D%20%3D%20160)