Answer:
Here we will find the pressure exerted by the gas?
Kinetic theory of gases shows the relationship between the Pressure of an ideal gas with the speed of molecules, or their average translational kinetic Energy
P
=0.667N/Vvr
where
P =Gas pressure?
vr = average translational kinetic energy of each atom =2.70×10⁻²³ J
N/V = number of gas atoms per volume = 6.76×10²⁰ atoms/cm³ = 6.76×10²⁶atoms/m³
P = 0.6667 x 6.76×10²⁶ x 2.70×10⁻²³ = 0.01217Pa
Differentiating both sides of PV = C with respect to t gives us ...
... P'V +PV' = 0
Filling in the given numbers gives us ...
... (40 kPa/min)(900 cm³) + V'(150 kPa) = 0
Solving for V' gives ...
... V' = -(40 kPa/min)(900 cm³)/(150 kPa)
.. V' = -240 cm³/min
The volume is decreasing at the rate of 240 cm³/min.
Answer:
Only the first graph with equation,
y = 11.5x
shows a proportional relationship between x and y.
Step-by-step explanation:
Only the first graph with equation,
y = 11.5x
shows a proportional relationship between x and y.
We know that y is proportional to x , written as,
y ∝ x
only when, y =
where k ≠ 0
The integers are 1 and 4.
1+4=5
1x4=4
Answer:
y = -2
Step-by-step explanation:
To find the equation of the tangent we apply implicit differentiation, and then we take apart dy/dx
The equation is
![y^2(y^2-4)=x^2(x^2-5)](https://tex.z-dn.net/?f=y%5E2%28y%5E2-4%29%3Dx%5E2%28x%5E2-5%29)
implicit differentiation give us
![\frac{d}{dx}[y^2(y^2-4)=x^2(x^2-5)]\\\\2y\frac{dy}{dx}(y^2-4)+y^2(2y\frac{dy}{dx})=2x(x^2-5)+x^2(2x)\\\\4y^3\frac{dy}{dx}-8y\frac{dy}{dx}=2x^3-10x+2x^3\\\\\frac{dy}{dx}=\frac{4x^3-10x}{4y^3-8y}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5By%5E2%28y%5E2-4%29%3Dx%5E2%28x%5E2-5%29%5D%5C%5C%5C%5C2y%5Cfrac%7Bdy%7D%7Bdx%7D%28y%5E2-4%29%2By%5E2%282y%5Cfrac%7Bdy%7D%7Bdx%7D%29%3D2x%28x%5E2-5%29%2Bx%5E2%282x%29%5C%5C%5C%5C4y%5E3%5Cfrac%7Bdy%7D%7Bdx%7D-8y%5Cfrac%7Bdy%7D%7Bdx%7D%3D2x%5E3-10x%2B2x%5E3%5C%5C%5C%5C%5Cfrac%7Bdy%7D%7Bdx%7D%3D%5Cfrac%7B4x%5E3-10x%7D%7B4y%5E3-8y%7D)
But we know that
![m=\frac{dy}{dx}\\y=mx+b](https://tex.z-dn.net/?f=m%3D%5Cfrac%7Bdy%7D%7Bdx%7D%5C%5Cy%3Dmx%2Bb)
Hence, for the point (0,-2) and by replacing for dy/dx
![m=\frac{dy}{dx}_{(0,-2)}=\frac{4(0)+10(0)}{4(-2)^3-8(-2)}=0](https://tex.z-dn.net/?f=m%3D%5Cfrac%7Bdy%7D%7Bdx%7D_%7B%280%2C-2%29%7D%3D%5Cfrac%7B4%280%29%2B10%280%29%7D%7B4%28-2%29%5E3-8%28-2%29%7D%3D0)
Hence m=0, that is, the tangent line to the point is a horizontal line that cross the y axis for y=-2. The equation is:
y=(0)x+b = -2
HOPE THIS HELPS!!