Which survey is biased?

According to Charles's law, if the pressure is held constant, the volume of a gas varies directly with the temperature measure

Ksju [112]

Answer:

V₂ = 252 cm³

Step-by-step explanation:

Charles's law = If the pressure is held constant, the volume of a gas varies directly with the temperature measured on the Kelvin scale.

Mathematically, $V=kT$

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

We have, V₁ = 192 cm³, T₁ = 112 K, T₂ = 147 cm³, V₂ = ?

Using the above relation of Charles's law :

$V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{192\times 147}{112}\\\\V_2=252\ cm^3$

So, the new temperature is 252 cm³.

7 0

3 years ago

Simplify 16.55(8) + 5.97(3)

34kurt

Answer:

150.31

Step-by-step explanation:

First, use Order of Operation to multiply.

16.55(8) + 5.97(3)

= 132.4 + 17.91

= 150.31

7 0

3 years ago

Read 2 more answers

3y''-6y'+6y=e*x sexcx

Simora [160]

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

$3r^2-6r+6=0\iff r^2-2r+2=0$

which has roots at $r=1\pm i$ . This admits the two fundamental solutions

$y_1=e^x\cos x$
$y_2=e^x\sin x$

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

$u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx$
$u_2=\dfrac13x$

Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$