Ask question

Ask question

All categories

3 years ago

10

HELP!!! i’ve been stuck at this question for a whole day

2 answers:

UkoKoshka [18]3 years ago

8 0

Have you done it already and when is this due

Nookie1986 [14]3 years ago

8 0

I’m stuck on algebra if anyone good help ASAP

You might be interested in

What is meant by the phrase " a consistent method of measurement "? someone help plss

Mars2501 [29]

<span><span>D.</span><span>Measurements are taken in a way that is the same every time.- apex

</span></span>

8 0

3 years ago

Read 2 more answers

What do you understand by family health?<br>

AleksandrR [38]

Answer:

Family health provides, besides the cure of disease, various activities and programs in prevention, rehabilitation, and health promotion for all members of families/households. pls mark me branilest

3 0

3 years ago

Read 2 more answers

Explain how the earth is round and not flat

frosja888 [35]

It is a conglomeration of rock that collected together over time, and it rotating in a spinning ball that is orbiting a sun that is in the middle of our solar system.

You should look at a picture of the earth from a satellite.<span />

5 0

4 years ago

Read 2 more answers

The interior space of large box is kept at 30 C. The walls of the box are 3 m high and have a ‘sandwich’ construction consisting

White raven [17]

Answer:

$\frac{\dot Q}{A} =20.129\ W.m^{-2}$

$T_1=27.58\ ^{\circ}C$ & $T_2=2.41875\ ^{\circ}C$

Explanation:

Given:

interior temperature of box, $T_i=30^{\circ}C$

height of the walls of box, $h=3\ m$

thickness of each layer of bi-layered plywood, $x_p=1.25\ cm=0.0125\ m$

thermal conductivity of plywood, $k_p=0.104\ W.m^{-1}.K^{-1}$

thickness of sandwiched Styrofoam, $x_s=5\ cm=0.05\ m$

thermal conductivity of Styrofoam, $k_s=0.04\ W.m^{-1}.K^{-1}$

exterior temperature, $T_o=0^{\circ}C$

<u>From the Fourier's law of conduction:</u>

$\dot Q=\frac{dT}{(\frac{x}{kA}) }$

$\dot Q=\frac{dT}{R_{th} }$ ....................................(1)

<u>Now calculating the equivalent thermal resistance for conductivity using electrical analogy:</u>

$R_{th}=R_p+R_s+R_p$

$R_{th}=\frac{x_p}{k_p.A}+\frac{x_s}{k_s.A}+\frac{x_p}{k_p.A}$

$R_{th}=\frac{1}{A} (\frac{x_p}{k_p}+\frac{x_s}{k_s}+\frac{x_p}{k_p})$

$R_{th}=\frac{1}{A} (\frac{0.0125}{0.104}+\frac{0.05}{0.04}+\frac{0.0125}{0.104})$

$R_{th}=\frac{1.4904}{A}$ .....................(2)

Putting the value from (2) into (1):

$\dot Q=\frac{30-0}{\frac{1.4904}{A} }$

$\dot Q=\frac{30\ A}{1.4904}$

$\frac{\dot Q}{A} =20.129\ W.m^{-2}$ is the heat per unit area of the wall.

The heat flux remains constant because the area is constant.

<u>For plywood-Styrofoam interface from inside:</u>

$\frac{\dot Q}{A} =k_p.\frac{T_i-T_1}{x_p}$

$20.129=0.104\times \frac{30-T_1}{0.0125}$

$T_1=27.58\ ^{\circ}C$

&<u>For Styrofoam-plywood interface from inside:</u>

$\frac{\dot Q}{A} =k_s.\frac{T_1-T_2}{x_s}$

$20.129=0.04\times \frac{27.58-T_2}{0.05}$

$T_2=2.41875\ ^{\circ}C$

4 0

3 years ago

Visible light falls into wavelength ranges of 400-700 nm, for which 1 m = 1 × 10 9 nm . The energy and wavelength of light are r

Paul [167]

Answer:

E = 3.54 x 10⁻¹⁹ J

Explanation:

The energy of the photon can be given in terms of its wavelength by the use of the following formula:

$E = \frac{hc}{\lambda}$

where,

E = energy = ?

h = Plank's Constant = 6.626 x 10⁻³⁴ Js

c = speed of light = 2.998 x 10⁸ m/s

λ = wavelength of light = 560.6 nm = 5.606 x 10⁻⁷ m

Therefore,

$E = \frac{(6.626\ x\ 10^{-34}\ Js)(3\ x\ 10^8\ m/s)}{5.606\ x\ 10^{-7}\ m}\\$

<u>E = 3.54 x 10⁻¹⁹ J</u>

7 0

3 years ago