Answer:
Explanation:
An object falling loses gravitational potential energy and gains kinetic energy. The gravity potential is the gravitational potential energy per unit mass. This energy comes from the gravitational potential energy released when the water falls. ... At 0, all the energy is in gravitational potential energy.
15+3=18km/hour
Think about it like this. The boat is going 15 faster than the river, and the river is going 3 faster than the bank, so the boat is going 18 faster than the river bank
<span>Now that you know the time to reach its maximum height, you have enough information to find out the initial velocity of the second arrow. Here's what you know about it: its final velocity is 0 m/s (at the maximum height), its time to reach that is 2.8 seconds, but wait! it was fired 1.05 seconds later, so take off 1.05 seconds so that its time is 1.75 seconds, and of course gravity is still the same at -9.8 m/s^2. Plug those numbers into the kinematic equation (Vf=Vi+a*t, remember?) for 0=Vi+-9.8*1.75 and solve for Vi to get.......
17.15 m/s</span>
Answer:
![F \approx 19.5 N](https://tex.z-dn.net/?f=F%20%5Capprox%2019.5%20N)
Explanation:
From the question we are told that:
Pressure of ![P_{CO_2}=1.50 * 105 Pa.](https://tex.z-dn.net/?f=P_%7BCO_2%7D%3D1.50%20%2A%20105%20Pa.)
Bottle cap area ![A_b= 4.40 * 10-4 m^2](https://tex.z-dn.net/?f=A_b%3D%204.40%20%2A%2010-4%20m%5E2)
Generally the equation for Resultant pressure
is give as is mathematically given by
![P_r=P_{CO_2}-P_a](https://tex.z-dn.net/?f=P_r%3DP_%7BCO_2%7D-P_a)
Where
![P_a=atmospheric\ pressure = 1.013*10^5 pa](https://tex.z-dn.net/?f=P_a%3Datmospheric%5C%20pressure%20%3D%201.013%2A10%5E5%20pa)
![P_r=1.50 * 105 Pa-1.013*10^5 pa](https://tex.z-dn.net/?f=P_r%3D1.50%20%2A%20105%20Pa-1.013%2A10%5E5%20pa)
![P_r=0.487*10^5 pa](https://tex.z-dn.net/?f=P_r%3D0.487%2A10%5E5%20pa)
Generally the equation for Force exerted by screw F is give as is mathematically given by
![F = P*A\\F = 0.487*10^5*4.00*10^-4\\ F = 19.48 N](https://tex.z-dn.net/?f=F%20%3D%20P%2AA%5C%5CF%20%3D%200.487%2A10%5E5%2A4.00%2A10%5E-4%5C%5C%20F%20%3D%2019.48%20N)
![F \approx 19.5 N](https://tex.z-dn.net/?f=F%20%5Capprox%2019.5%20N)