Answer:
(a) Velocity will be 17.146 m/sec
(b) Volume flow rate will be ![66.871\times 10^{-3}m^3/sec](https://tex.z-dn.net/?f=66.871%5Ctimes%2010%5E%7B-3%7Dm%5E3%2Fsec)
Explanation:
We have given height of the dam h = 15 m
Acceleration due to gravity ![g=9.8m/sec^2](https://tex.z-dn.net/?f=g%3D9.8m%2Fsec%5E2)
Effective area ![A=3.90\times 10^{-3}m^2](https://tex.z-dn.net/?f=A%3D3.90%5Ctimes%2010%5E%7B-3%7Dm%5E2)
(a) From conservation of energy
Potential energy at the top will be equal to kinetic energy at the bottom
So ![mgh=\frac{1}{2}mv^2](https://tex.z-dn.net/?f=mgh%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
![v=\sqrt{2gh}=\sqrt{2\times 9.8\times 15}=17.146m/sec](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B2gh%7D%3D%5Csqrt%7B2%5Ctimes%209.8%5Ctimes%2015%7D%3D17.146m%2Fsec)
(b) We have to find the volume flow rate
Volume flow rate is given by
![V=area\times velocity = 3.9\times 10^{-3}\times 17.146=66.871\times 10^{-3}m^3/sec](https://tex.z-dn.net/?f=V%3Darea%5Ctimes%20velocity%20%3D%203.9%5Ctimes%2010%5E%7B-3%7D%5Ctimes%2017.146%3D66.871%5Ctimes%2010%5E%7B-3%7Dm%5E3%2Fsec)
Answer:
We assume the second train was standing still and that momentum is conserved.
Then the product of mass and velocity before the collision is
(5000 kg)·(100 m/s) = 500,000 kg·m/s.
After the collision, where M is the mass of the second train, the momentum is
((5000+M) kg)·(50 m/s) = 500,000 kg·m/s
Dividing by 50 m/s and subtracting 5000 kg, we have
(5000 +M) kg = 10,000 kg
M kg = 5000 kg
The mass of the second train is 5000 kg
B) 200 km/h
Each car is traveling at 100 km/h, so you add the speeds to gether since they are traviling in opposite directions.
For the purpose of the exercise, we can assume that the Earth is at perihelion (closest point to the Sun) on December 21st (in reality, it happens around January 4th) and that the Earth is at aphelion (farthest point from the Sun) on June 21st (in reality, this happens around July 4th). The distance Earth-Sun is the following:
- Perihelion: 147.1 milion km
- Aphelion: 152.1 milion km
- Average distance: 149.6 milion km
So, we can calculate the percentage change with respect to the average distance as:
Answer:
-0.01N
Explanation:
I'm assuming you're asking for the net force of the paper.
0.01N - 0.02N = -0.01N