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3 years ago

15

Sarah needs to create an architectural drawing for a museum building with an inclined surface. Which presentation view will be t

he best to help the client visualize the design?

1 answer:

prohojiy [21]3 years ago

7 0

Answer: auxiliary

Explanation: got it right

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Air is compressed slowly in a piston-cylinder assembly from an initial state where P1 = 1.4 bar, V1= 4.25 m^3, to a final state

lord [1]

Answer:

W=-940.36 KJ

Explanation:

Given that

$P_1=1\ bar,V_1=4.25 {m^3}$

$P_2=6.8\ bar$

Process follows pv=constant

So this is the isothermal process and work in isothermal process given as

$W=P_1V_1\ln \dfrac{P_1}{P_2}$

Now by putting the values (1.4 bar =140 KPa)

$W=P_1V_1\ln \dfrac{P_1}{P_2}$

$W=140\times 4.25 \ln \dfrac{1.4}{6.8}$

W=-940.36 KJ

Negative sign indicates that this is a compression process and work will given to the system.

7 0

4 years ago

Determine the magnitude of force P needed to start towing the 40kg crate.Also determine the location of the resultant normal for

erastova [34]

The distance from Point A=500 mm

Explanation:

M := 40kg c := 200mm

μs := 0.3 d := 3

a := 400mm e := 4

b := 800mm

Initial guesses: $N_{c}$ := 200 N P := 50N

Plz refer to the image below

5 0

3 years ago

If you know that the change in entropy of a cup of coffee where heat was added is 20 J/K, and that the temperature of the coffee

Anna35 [415]

Answer:

5000J

Explanation:

Given in the question that

Heat added to the coffee cup is, ΔS = 20 J/K

The temperature of the coffee, T = 250 K

Now, using the formula for the entropy change

$\bigtriangleup S=-\frac{\bigtriangleup H}{T}$ ...........(1)

Where,

ΔS is the entropy change

ΔH is the enthalpy change

T is the temperature of the system

substituting the values in the equation (1)

we get

$20=-\frac{\bigtriangleup H}{250}$

ΔH=250×20

ΔH=5000 J

6 0

3 years ago

Place test tubes containing approximately equal volumes in opposite positions on the rotor. Wait until all positions on the roto

Anvisha [2.4K]

Answer:

do it

Explanation:

they are giving you step by step instructions to do. i woul hate to have to work with you out anywhere in the world if you have to ask this. i am 15, and i would at least read the d a m n thing before i post it.

3 0

3 years ago

Liquid flows at steady state at a rate of 2 lb/s through a pump, which operates to raise the elevation of the liquid 100 ft from

Greeley [361]

Answer:

D) 1.04 Btu/s from the liquid to the surroundings.

Explanation:

Given that:

flow rate (m) = 2 lb/s

liquid specific enthalpy at the inlet ( $h_{1}=40.09$ Btu/lb)

liquid specific enthalpy at the exit ( $h_{2}=40.94$ Btu/lb)

initial elevation ( $z_1=0ft$ )

final elevation ( $z_2=100ft$ )

acceleration due to gravity (g) = 32.174 ft/s²

$W_{cv}$ = 3 Btu/s

The energy balance equation is given as:

$Q_{cv}-W{cv}+m[(h_1-h_2)+(\frac{V_1^2-V_2^2}{2})+g(z_1-z_2)]=0$

Since kinetic energy effects are negligible, the equation becomes:

$Q_{cv}-W{cv}+m[(h_1-h_2)+g(z_1-z_2)]=0$

Substituting values:

$Q_{cv}-(-3)+2[(40.09-40.94)+\frac{32.174(0-100)}{778*32.174} ]=0\\Q_{cv}+3+2[-0.85-0.1285 ]=0\\Q_{cv}+3+2(-0.9785)=0\\Q_{cv}+3-1.957=0\\Q_{cv}+1.04=0\\Q_{cv}=-1.04\\$

The heat transfer rate is 1.04 Btu/s from the liquid to the surroundings.

8 0

3 years ago