<span>The reason a static method can't access instance variable is because static references the class not a specific instance of the class so there is no instance variable to access.</span>
Answer:
In the air
Explanation:
There are three states of matter:
- Solids: in solids, the particles are tightly bond together by strong intermolecular forces, so they cannot move freely - they can only vibrate around their fixed position
- Liquids: in liquids, particles are more free to move, however there are still some intermolecular forces keeping them close to each other
- Gases: in gases, particles are completely free to move, as the intermolecular forces between them are negligible
For this reason, it is generally easier to compress/expand the volume of a gas with respect to the volume of a liquid.
In this problem, we are comparing water (which is a liquid) with air (which is a gas). From what we said above, this means that the change in volume is larger in the air rather than in the water.
Well i think the best answer would be A
The answer is 50% of the paper production
<span>Using the kinematic equations below, we can calculate the initial velocity required.
Angle of projectile = 60 degrees
Acceleration due to gravity (Ay) = -10 m/s^2 (negative because downward)
Height of projectile (Dy) = 2m
Vfy^2=Voy^2 +2*Ay*Dy
Vfy = 0 m/s because the vertical velocity slows to zero at the height of its trajection.
So... 0 = Voy^2 + 2(-10)(2)
0 = Voy^2 - 40
40 = Voy^2
Sqrt40 = Voy
6.32 m/s = Voy
THIS IS NOT THE ANSWER. THIS IS JUST THE INITIAL VELOCITY IN THE Y DIRECTION.
Using trigonometry, Tan 60 = Voy/Vox. Tan 60 = 6.32/Vox. Vox*Tan 60 = Vox
Vox = 10.95 m/s. Now, using Vox = 10.95 and Voy = 6.32, we can use pythagorean theorem to find the total Vo. A^2 +B^2 = C^2
10.95^2 + 6.32^2 = C^2
Solving for C = 12.64 m/s
This is the velocity required to hit the surface. You can also calculate a bunch of other stuff now using the other kinematic equations.
V = 12.64 m/s</span>
