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3 years ago

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How much heat is lost through a 3’× 5' single-pane window with a storm that is exposed to a 60°F temperature differential?A. 450

Btu/hB. 900 Btu/hC. 1350 Btu/hD. 1800 Btu/h

1 answer:

arsen [322]3 years ago

8 0

Answer:

A. 450 btu/h

Explanation:

We solve this problem by using this formula:

Q = U x TD x area

U = U value of used material

TD = Temperature difference = 60°

Q = heat loss

Area = 3x5 = 15

We first find U

R = 1/u

2 = 1/U

U = 1/2 = 0.5

Then when we put these values into the formula above, we would have:

Q = 0.5 x 15 x 60

Q = 450Btu/h

Therefore 450btu/h is the answer

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A stream of moist air flows into an air conditioner with an initial humidity ratio of 0.6 kg(vapor)/kg(dry air), and a dry air f

BaLLatris [955]

Answer:

$\omega_{out} = 0.867\,\frac{kg\,H_{2}O}{kg\,DA}$

Explanation:

The final humidity ratio is computed by the Principle of Mass Conservation:

Dry Air

$\dot m_{in} = \dot m_{out}$

Moist

$\dot m_{in} \cdot \omega_{in} + \dot m_{w} = \dot m_{out}\cdot \omega_{out}$

Then, the final humidity ratio is:

$\omega_{out} = \frac{\dot m_{in}\cdot \omega_{in}+\dot m_{w}}{\dot m_{out}}$

$\omega_{out} = \omega_{in} + \frac{\dot m_{w}}{\dot m_{out}}$

$\omega_{out} = 0.6\,\frac{kg\,H_{2}O}{kg\,DA} + \frac{0.4\,\frac{kg\,H_{2}O}{s} }{1.5\,\frac{kg\,DA}{s} }$

$\omega_{out} = 0.867\,\frac{kg\,H_{2}O}{kg\,DA}$

4 0

4 years ago

Write a program that randomly chooses between three different colors for displaying text on the screen. Use a loop to display tw

11Alexandr11 [23.1K]

Answer:

INCLUDE Irvine32.inc

.data

msgIntro byte "This is Your Name's fourth assembly extra credit program. Will randomly",0dh,0ah

byte "choose between three different colors for displaying twenty lines of text,",0dh,0ah

byte "each with a randomly chosen color. The color probabilities are as follows:",0dh,0ah

byte "White=30%,Blue=10%,Green=60%.",0dh,0ah,0

msgOutput byte "Text printed with one of 3 randomly chosen colors",0

.code

main PROC

;

//Intro Message

mov edx,OFFSET msgIntro ;intro message into edx

call WriteString ;display msgIntro

call Crlf ;endl

call WaitMsg ;pause message

call Clrscr ;clear screen

call Randomize ;seed the random number generator

mov edx, OFFSET msgOutput;line of text

mov ecx, 20 ;counter (lines of text)

L1:;//(Loop - Display Text 20 Times)

call setRanColor ;calls random color procedure

call SetTextColor ;calls the SetTextColor from library

call WriteString ;display line of text

call Crlf ;endl

loop L1

exit

main ENDP

;--

setRanColor PROC

;

; Selects a color with the following probabilities:

; White = 30%, Blue = 10%, Green = 60%.

; Receives: nothing

; Returns: EAX = color chosen

;--

mov eax, 10 ;range of random numbers (0-9)

call RandomRange ;EAX = Random Number

.IF eax >= 4 ;if number is 4-9 (60%)

mov eax, green ;set text green

.ELSEIF eax == 3 ;if number is 3 (10%)

mov eax, blue ;set text blue

.ELSE ;number is 0-2 (30%)

mov eax, white ;set text white

.ENDIF ;end statement

ret

setRanColor ENDP

6 0

4 years ago

On Beverly's last project, the team identified only a few lessons learned. Which approach to lessons learned

Afina-wow [57]

Option C (making lessons learned a regular part of meetings) is the correct approach.

As nothing more than a general rule, typically construction companies only plan lessons that have been learned exercises or initiatives towards the end of a particular endeavor or segment.
As almost a result of the team knowing, valuable lessons are intended to increase the comprehensive implementation of quality management practices as well as deadlines.

Aside from this, none of the choices are viable methods to learning lessons or gaining knowledge from the past. As a result, the methodology outlined above is the appropriate one.

Learn more about project teamwork here:

brainly.com/question/14279121

7 0

3 years ago

A pin must be inserted into a collar of the same steel using an expansion fit. The coefficient of thermal expansion of the metal

nirvana33 [79]

Answer:

a) the temperature to which the pin must be cooled for assembly is $T_2 = -101.89^ \ ^0}C$

b) the radial pressure at room temperature after assembly is $P_f = 62.8 \ MPa$

c) the safety factor in the resulting assembly = 6.4

Explanation:

Coefficient of thermal expansion $\alpha = 12.3*10^{-6} \ ^0 C$

Yield strength $\sigma_y$ = 400 MPa

Modulus of elasticity (E) = 209 GPa

Room Temperature $T_1$ = 20°C

outer diameter of the collar $D_o = 95 \ mm$

inner diameter of the collar $D_i = 60 \ mm$

pin diameter $D_p$ = $60.03 \ mm$

Clearance c = 0.06 mm

a)

The temperature to which the pin must be cooled for assembly can be calculated by using the formula:

$(D_i - c )-D_p = \alpha * D_p(T_2-T_1)$

$(60-0.06)-60.03=12.3*10^{-6}*60.03(T_{2}-20^{0}C)$

-0.09 = $7.38369*10^{-4}$ $(T_{2}-20^{0}C)$

-0.09 = $7.38369*10^{-4}T_2$ $\ \ - \ \ 0.01476738$

$-0.09 + 0.01476738$ = $7.38369*10^{-4}T_2$

−0.07523262 = $7.38369*10^{-4}T_2$

$T_2 = \frac{-0.07523262}{7.38369*10^{-4}}$

$T_2 = -101.89^ \ ^0}C$

b)

To determine the radial pressure at room temperature after assembly ;we have:

$P_f = \frac{E * (D_p-D_i)(D_o^2-D_1^2)}{D_i*D_o} \\ \\ \\ P_f = \frac{209*10^9* 0.03(95^2-60^2)}{60*95^2} \\ \\ P_f = 62815789.47 \ Pa \\ \\ P_f = 62.8 \ MPa$

c) the safety factor of the resulting assembly is calculated as:

safety factor = $\frac{Yield \ strength }{walking \ stress}$

safety factor = $\frac{400}{62.8}$

safety factor = 6.4

Thus, the safety factor in the resulting assembly = 6.4

4 0

4 years ago

This test should be performed on all cord sets, receptacles that aren't part of a building or structure's permanent wiring, and

vova2212 [387]

Answer:

A continuity test

Explanation:

A continuity test is used to verified that current will flow in an electrical circuit, it performed by placing a small voltage across the chosen path. continuity test ensure that the equipment grounding conductor is electrically continuous and this test is perform on all the cord sets, receptacles that aren't part of a building or structure's permanent wiring, and cord-and-plug connected equipment required to be grounded. example of equipment used in testing current flow in continuity test are Analog multi-meter, voltage/continuity tester etc.

Continuity test and terminal connection test are the two test required by OSHA on all electrical equipment

8 0

3 years ago