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irina [24]
2 years ago
10

The gas law for a fixed mass m of an ideal gas at absolute temperature T, pressure P, and volume V is PV=mRT, where R is the gas

constant. Find the partial derivatives
Physics
1 answer:
Solnce55 [7]2 years ago
4 0

Answer:

the equation is

P*V = mRT.

The partial derivatives are:

For P = mRT/V.

dP/dV = -\frac{mRT}{V^2}

dP/dT = \frac{mR}{V}

now, if we take:

V = mRT/P

dV/dT = \frac{mR}{P}

dV/dP = -\frac{mRT}{P^2} = (dP/dV)^{-1} = -\frac{V^2}{mRT}

If we take T = PV/mR

dT/dP = \frac{V}{mR} = (dP/dT)^{-1}

dT/dV = \frac{P}{mR} = (dV/dT)^{-1}

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The force that gravitation exerts upon a body, equal to the mass of the body times the local acceleration of gravity is known as weight.

Weight is the force of gravity acting on a body.

The formula is :

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Here,

W is the weight or force acting on the body. m is the mass of the body, and g is the gravitational acceleration.

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1 year ago
If I was a scientist and I wanted to measure the intensity of an earthquake I would use
g100num [7]

Answer:

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Explanation:

6 0
2 years ago
Help! <br><br> I need to do this quickly!!!
umka2103 [35]
D is the answer. It is a firm statement.
8 0
2 years ago
A 9-μC positive point charge is located at the origin and a 6 μC positive point charge is located at x = 0.00 m, y = 1.0 m. Find
sukhopar [10]

Answer:

The coordinates of the point is (0,0.55).

Explanation:

Given that,

First charge q_{1}=9\times10^{-6}\ C at origin

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Second charge at point P = (0,1)

We assume that,

The net electric field between the charges is zero at mid point.

Using formula of electric field

E=\dfrac{kq}{r^2}

0=\dfrac{k\times9\times10^{-6}}{d^2}+\dfrac{k\times6\times10^{-6}}{(1-d)^2}

\dfrac{(1-d)}{d}=\sqrt{\dfrac{6}{9}}

\dfrac{1}{d}=\dfrac{\sqrt{6}}{3}+1

\dfrac{1}{d}=1.82

d=\dfrac{1}{1.82}

d=0.55\ m

Hence, The coordinates of the point is (0,0.55).

3 0
3 years ago
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Hope this helps!!
7 0
3 years ago
Read 2 more answers
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