Please answer the following question down below. (25 points)

The root mean square velocity of one mole of a monoatomic gas having molar mass M is U rms . The relation between average kineti

garik1379 [7]

root means square velocity of one mole of a monoatomic gas having molar mas*s m(M) is Urms. The relation between average kinetic energy (e)of the gas and Urms is Urms=M2E

StarrySoul please check it...

8 0

2 years ago

4. From 5 faculty members, a committee of 2 is to be formed. In how many ways can this be done?

ohaa [14]

Answer:

3

Step-by-step explanation:

7 0

3 years ago

A large corporation starts at time t = 0 to invest part of its receipts continuously at a rate of P dollars per year in a fund f

Andrews [41]

Answer:

$A = \frac{P}{r}\left( e^{rt} -1 \right)$

Step-by-step explanation:

This is a separable differential equation. Rearranging terms in the equation gives

$\frac{dA}{rA+P} = dt$

Integration on both sides gives

$\int \frac{dA}{rA+P} = \int dt$

where $c$ is a constant of integration.

The steps for solving the integral on the right hand side are presented below.

$\int \frac{dA}{rA+P} = \begin{vmatrix} rA+P = m \implies rdA = dm\end{vmatrix} \\\\\phantom{\int \frac{dA}{rA+P} } = \int \frac{1}{m} \frac{1}{r} \, dm \\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \int \frac{1}{m} \, dm\\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |m| + c \\\\&\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |rA+P| +c$

Therefore,

$\frac{1}{r} \ln |rA+P| = t+c$

Multiply both sides by $r.$

$\ln |rA+P| = rt+c_1, \quad c_1 := rc$

By taking exponents, we obtain

$e^{\ln |rA+P|} = e^{rt+c_1} \implies |rA+P| = e^{rt} \cdot e^{c_1} rA+P = Ce^{rt}, \quad C:= \pm e^{c_1}$

Isolate $A$ .

$rA+P = Ce^{rt} \implies rA = Ce^{rt} - P \implies A = \frac{C}{r}e^{rt} - \frac{P}{r}$

Since $A = 0$ when $t=0$ , we obtain an initial condition $A(0) = 0$ .

We can use it to find the numeric value of the constant $c$ .

Substituting $0$ for $A$ and $t$ in the equation gives

$0 = \frac{C}{r}e^{0} - \frac{P}{r} \implies \frac{P}{r} = \frac{C}{r} \implies C=P$

Therefore, the solution of the given differential equation is

$A = \frac{P}{r}e^{rt} - \frac{P}{r} = \frac{P}{r}\left( e^{rt} -1 \right)$

4 0

3 years ago

at best buy they have a 60"tv that sells for $1999 and is on sale for 20% off. the sales tax is 8.8%. what is the final cost

sweet [91]

I believe the answer is $1,458.47

6 0

3 years ago

The measured weights of 1,000 men in a certain village follow a normal distribution with a mean of 150 pounds and a standard dev

Hitman42 [59]

Z = (X-Mean)/SD
z1 = (165 - 150)/15 = +1 
z2 = (135 - 150)/15 = - 1 
According to the Empirical Rule 68-95-99.7 
Mean +/- 1SD covers 68% of the values 
100% - 68% = 32% 
The remaining 32% is equally distributed below z = - 1 and z = +1 
32%/2 = 16%

Therefore,

a) Number of men weighing more than 165 pounds = 16% of 1000 = 160 
b) Number of men weighing less than 135 pounds = 16% of 1000 = 160

3 0

3 years ago

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