All categories

3 years ago

Which of these materials are insulators? Select the THREE (3) that apply.

Engineering

1 answer:

lubasha [3.4K]3 years ago

5 0

Answer:

wool, rubber, and plastic

Explanation:

You might be interested in

An analog baseband audio signal with a bandwidth of 4kHz is transmitted through a transmission channel with additive white noise

siniylev [52]

Answer:

2k20

Explanation:

4k ✈

8 0

3 years ago

Omplete the following program: [0.5 X 4 = 2]

borishaifa [10]

I’m confused on this one

4 0

3 years ago

The top surface of an L = 5mmthick anodized aluminum plate is irradiated with G = 1000 W/m2 while being simultaneously exposed

puteri [66]

Answer:

$J=1963W/m^2$

$q_{rad}=963w/m^2$

$\triangle T= -0.378k/s$

Explanation:

From the question we are told that:

$L=5mm => 5*10^{-3}\\Irradiation G=1000W/m^2\\h=50W/m^2\\T_{infinity} = 25C.\\Plate\ temperature\ T_p= 400 K\\\alpha=0.14\\E=0.76$

at Temp=400K

$E=2702kg/m^2,c=949J/kgk$

Generally the equation for Radiosity is mathematically given by

$J=eG+\in E_p$

$J=(1-\alpha)G+\in \sigma T^4$

$J=(1-0.14)1000+0.76 (5.67*10^_{8}) (400)^4$

$J=1963W/m^2$

Generally the equation for net radiation heat flux $q_{rad}$ is mathematically given by

$q_{rad}=J-G\\q_{rad}=1963-1000$

$q_{rad}=963w/m^2$

Generally the equation for and the rate of plate temp $\triangleT$ is mathematically given by

$\triangle T= -\frac{q_{con} +q_{rad}}{Ecl}$

$\triangle T= \frac{45(400)-(30+273+963)}{(2702*949*0.005)}$

$\triangle T= -0.378k/s$

6 0

3 years ago

The compressor in a refrigerator compresses saturated R-134a vapor at 0°F to 200 psia. Calculate the work required by this compr

kvasek [131]

Answer:

The work required is W = 20.2 BTU per lbm

Explanation:

The value of entropy & enthalpy at initial conditions are

$h_{1}$ = 103.1

S = 0.225

Final enthalpy

$h_{2}$ = 123.3

Therefore work done

W = $h_{1}$ - $h_{2}$

W = 103.1 - 123.3

W = - 20.2 BTU per lbm

Therefore the work required is W = 20.2 BTU per lbm

6 0

3 years ago

I am making composites of silicone rubber and copper particles; by mixing thermally conductive particles into the thermally insu

Gelneren [198K]

Answer:

The composition of Composite of Volume basis is 2.154% Copper and 97.83% Rubber.

Explanation:

Let us assume that the total mass of composite is 100 lbm So as per the given conditions

15 lbm is copper and 85 lbm is rubber.
Density of rubber is 70 lbm/ft3
Specific gravity of Copper is 9

So As per the formula of specific gravity

$S_{cu}=\frac{\rho_{cu}}{\rho_w}$

here density of water is 62.4 lbm/ft3

Solving for Density of Copper gives

$S_{cu}=\frac{\rho_{cu}}{\rho_w}\\9=\frac{\rho_{cu}}{62.4}\\\rho_{cu}=9 \times 62.4\\\rho_{cu}=561.5 lbm/ft3$

For composition on volume basis, volume of individual components and composite are calculated as

$V_{cu}=\frac{m_{cu}}{\rho_{cu}}\\V_{cu}=\frac{15}{561.5}\\V_{cu}=0.0267 ft^3\\\\V_{r}=\frac{m_{r}}{\rho_{r}}\\V_{r}=\frac{85}{70}\\V_{r}=1.214 ft^3\\\\V_{c}=V_{r}+V_{cu}\\V_{c}=1.214+0.0267 \\V_{c}=1.2409 ft^3$

The composition is given as

$c_{cu}=\frac{V_{cu}}{V_{c}}\\c_{cu}=\frac{0.0267}{1.2409} \times 100 \%\\c_{cu}=2.154 \%\\\\c_{r}=\frac{V_{r}}{V_{c}}\\c_{r}=\frac{1.214}{1.2409} \times 100 \%\\c_{r}=97.83 \%$

So the composition of Composite of Volume basis is 2.154% Copper and 97.83% Rubber.

6 0

3 years ago