Recall this gas law:
= ![\frac{P₂V}{T₂}](https://tex.z-dn.net/?f=%5Cfrac%7BP%E2%82%82V%7D%7BT%E2%82%82%7D)
P₁ and P₂ are the initial and final pressures.
V₁ and V₂ are the initial and final volumes.
T₁ and T₂ are the initial and final temperatures.
Given values:
P₁ = 475kPa
V₁ = 4m³, V₂ = 6.5m³
T₁ = 290K, T₂ = 277K
Substitute the terms in the equation with the given values and solve for Pf:
![\frac{475*4}{290} = \frac{P₂*6.5}{277}](https://tex.z-dn.net/?f=%5Cfrac%7B475%2A4%7D%7B290%7D%20%3D%20%5Cfrac%7BP%E2%82%82%2A6.5%7D%7B277%7D)
<h3>P₂ = 279.2kPa</h3>
Answer:
995 N
Explanation:
Weight of surface, w= 4000N
Gravitational constant, g, is taken as 9.81 hence mass, m of surface is W/g where W is weight of surface
m= 4000/9.81= 407.7472
Using radius of orbit of 6371km
The force of gravity of satellite in its orbit, ![F=\frac {GMm}{(2r)^{2}}=\frac {GMm}{4(r)^{2}}](https://tex.z-dn.net/?f=F%3D%5Cfrac%20%7BGMm%7D%7B%282r%29%5E%7B2%7D%7D%3D%5Cfrac%20%7BGMm%7D%7B4%28r%29%5E%7B2%7D%7D)
Where
and ![M=5.94*10^{24}](https://tex.z-dn.net/?f=M%3D5.94%2A10%5E%7B24%7D)
![F=\frac {(6.67*10^{-11}*5.94*10^{24}*407.7472)}{4*({6.371*10^{6}m)}^{2}}](https://tex.z-dn.net/?f=F%3D%5Cfrac%20%7B%286.67%2A10%5E%7B-11%7D%2A5.94%2A10%5E%7B24%7D%2A407.7472%29%7D%7B4%2A%28%7B6.371%2A10%5E%7B6%7Dm%29%7D%5E%7B2%7D%7D)
F= 995.01142 then rounded off
F=995N
Answer:
![t=12.25\ seconds](https://tex.z-dn.net/?f=t%3D12.25%5C%20seconds)
Explanation:
<u>Diagonal Launch
</u>
It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.
The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is
![x=v_ocos\theta t](https://tex.z-dn.net/?f=x%3Dv_ocos%5Ctheta%20t)
![\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3Dy_o%2Bv_osin%5Ctheta%20%5C%20t-%5Cfrac%7Bgt%5E2%7D%7B2%7D)
Where vo is the magnitude of the initial velocity,
is the angle, t is the time and g is the acceleration of gravity
The maximum height the object can reach can be computed as
![\displaystyle t=\frac{v_osin\theta}{g}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20t%3D%5Cfrac%7Bv_osin%5Ctheta%7D%7Bg%7D)
There are two times where the value of y is
when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making ![y=y_o](https://tex.z-dn.net/?f=y%3Dy_o)
![\displaystyle y_o=y_o+v_osin\theta\ t-\frac{gt^2}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y_o%3Dy_o%2Bv_osin%5Ctheta%5C%20t-%5Cfrac%7Bgt%5E2%7D%7B2%7D)
Removing
and dividing by t (t different of zero)
![\displaystyle 0=v_osin\theta-\frac{gt}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%200%3Dv_osin%5Ctheta-%5Cfrac%7Bgt%7D%7B2%7D)
Then we find the total flight as
![\displaystyle t=\frac{2v_osin\theta}{g}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20t%3D%5Cfrac%7B2v_osin%5Ctheta%7D%7Bg%7D)
We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is
![\boxed{t=24.5/2=12.25\ seconds}](https://tex.z-dn.net/?f=%5Cboxed%7Bt%3D24.5%2F2%3D12.25%5C%20seconds%7D)
<span>While you're going to the store, your acceleration changes. Some times it increases your overall speed sometimes it reduces it. Constant acceleration does not occur because it would mean that you would constantly accelerate and eventually go past the store. Even reduction of speed is a type of acceleration in physics. When you reach it, we can then calculate how much your velocity was on average and analyze how changing acceleration would've affected it.</span>
Answer:
t=6
Explanation:
Multiply to remove the variable from the denominator.