Answer:
a)T total = 2*Voy/(g*sin( α ))
b)α = 0º , T total≅∞ (the particle, goes away horizontally indefinitely)
α = 90º, T total=2*Voy/g
Explanation:
Voy=Vo*sinα
- Time to reach the maximal height :
Kinematics equation: Vfy=Voy-at
a=g*sinα ; g is gravity
if Vfy=0 ⇒ t=T ; time to reach the maximal height
so:
0=Voy-g*sin( α )*T
T=Voy/(g*sin( α ))
- Time required to return to the starting point:
After the object reaches its maximum height, the object descends to the starting point, the time it descends is the same as the time it rises.
So T total= 2T = 2*Voy/(g*sin( α ))
The particle goes totally horizontal, goes away indefinitely
T total= 2*Voy/(g*sin( α )) ≅∞
T total=2*Voy/g
Rigidbodies are components that allow a GameObject<u> to react to real-time physics. </u>
Explanation:
- Rigidbodies are components that allow a GameObject to react to real-time physics. This includes reactions to forces and gravity, mass, drag and momentum. You can attach a Rigidbody to your GameObject by simply clicking on Add Component and typing in Rigidbody2D in the search field.
- A rigidbody is a property, which, when added to any object, allows it to interact with a lot of fundamental physics behaviour, like forces and acceleration. You use rigidbodies on anything that you want to have mass in your game.
- You can indeed have a collider with no rigidbody. If there's no rigidbody then Unity assumes the object is static, non-moving.
- If you had a game with only two objects in it, and both move kinematically, in theory you would only need a rigidbody on one of them, even though they both move.
Answer:
Acceleration and velocity Newton's second law says that when a constant force acts on a massive body, it causes it to accelerate, i.e., to change its velocity, at a constant rate. In the simplest case, a force applied to an object at rest causes it to accelerate in the direction of the force.
Answer:
T'=92.70°C
Explanation:
To find the temperature of the gas you use the equation for ideal gases:

V: volume = 3000cm^3 = 3L
P: pressure = 1250mmHg; 1 mmHg = 0.001315 atm
n: number of moles
R: ideal gas constant = 0.082 atm.L/mol.K
T: temperature = 27°C = 300.15K
For the given values you firs calculate the number n of moles:
![n=\frac{PV}{RT}=\frac{(1520[0.001315atm])(3L)}{(0.082\frac{atm.L}{mol.K})(300.15K)}=0.200moles](https://tex.z-dn.net/?f=n%3D%5Cfrac%7BPV%7D%7BRT%7D%3D%5Cfrac%7B%281520%5B0.001315atm%5D%29%283L%29%7D%7B%280.082%5Cfrac%7Batm.L%7D%7Bmol.K%7D%29%28300.15K%29%7D%3D0.200moles)
this values of moles must conserve when the other parameter change. Hence, you have V'=2L and P'=3atm. The new temperature is given by:

hence, T'=92.70°C