Check the level of motor oil in your engine by ?

An excavation is at risk for cave-in and water accumulation because of the excess soil that has accumulated. What type of excava

N76 [4]

Answer:Topographic map. Contour line. Learning Objectives. After completing this chapter, you will be able to: □ Define civil engineering and civil drafting.

Explanation:

4 0

3 years ago

Read 2 more answers

1. A glass window of width W = 1 m and height H = 2 m is 5 mm thick and has a thermal conductivity of kg = 1.4 W/m*K. If the inn

emmasim [6.3K]

Answer:

1. $\dot Q=19600\ W$

2. $\dot Q=120\ W$

Explanation:

1.

Given:

height of the window pane, $h=2\ m$

width of the window pane, $w=1\ m$

thickness of the pane, $t=5\ mm= 0.005\ m$

thermal conductivity of the glass pane, $k_g=1.4\ W.m^{-1}.K^{-1}$

temperature of the inner surface, $T_i=15^{\circ}C$

temperature of the outer surface, $T_o=-20^{\circ}C$

<u>According to the Fourier's law the rate of heat transfer is given as:</u>

$\dot Q=k_g.A.\frac{dT}{dx}$

here:

A = area through which the heat transfer occurs = $2\times 1=2\ m^2$

dT = temperature difference across the thickness of the surface = $35^{\circ}C$

dx = t = thickness normal to the surface = $0.005\ m$

$\dot Q=1.4\times 2\times \frac{35}{0.005}$

$\dot Q=19600\ W$

2.

air spacing between two glass panes, $dx=0.01\ m$

area of each glass pane, $A=2\times 1=2\ m^2$

thermal conductivity of air, $k_a=0.024\ W.m^{-1}.K^{-1}$

temperature difference between the surfaces, $dT=25^{\circ}C$

<u>Assuming layered transfer of heat through the air and the air between the glasses is always still:</u>

$\dot Q=k_a.A.\frac{dT}{dx}$

$\dot Q=0.024\times 2\times \frac{25}{0.01}$

$\dot Q=120\ W$

5 0

3 years ago

A missile flying at high speed has a stagnation pressure and temperature of 5 atm and 598.59 °R respectively. What is the densit

alexdok [17]

Answer:

$5.31\frac{kg}{m^3}$

Explanation:

Approximately, we can use the ideal gas law, below we see how we can deduce the density from general gas equation. To do this, remember that the number of moles n is equal to $\frac{m}{M}$ , where m is the mass and M the molar mass of the gas, and the density is $\frac{m}{V}$ .

For air $M=28.66*10^{-3}\frac{kg}{mol}$ and $\frac{5}{9}R=K$

So, $598.59 R*\frac{5}{9}=332.55K$

$pV=nRT\\pV=\frac{m}{M}RT\\\frac{m}{V}=\frac{pM}{RT}\\\rho=\frac{pM}{RT}\\\rho=\frac{(5atm)28.66*10^{-3}\frac{kg}{mol}}{(8.20*10^{-5}\frac{m^3*atm}{K*mol})332.55K}=5.31\frac{kg}{m^3}$