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3 years ago

Is the gap store an example of commercial construction? Yes or no

Engineering

1 answer:

user100 [1]3 years ago

5 0

Answer:

Explanation:

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A welding rod with κ = 30 (Btu/hr)/(ft ⋅ °F) is 20 cm long and has a diameter of 4 mm. The two ends of the rod are held at 500 °

SOVA2 [1]

Answer:

In Btu:

Q=0.001390 Btu.

In Joule:

Q=1.467 J

Part B:

Temperature at midpoint=274.866 C

Explanation:

Thermal Conductivity=k=30 (Btu/hr)/(ft ⋅ °F)= $\frac{30}{3600} (Btu/s)/(ft.F)=8.33*10^{-3} (Btu/s)/(ft.F)$

Thermal Conductivity is SI units:

$k=30(Btu/hr)/(ft.F) * \frac{1055.06}{3600*0.3048*0.556} \\k=51.88 W/m.K$

Length=20 cm=0.2 m= (20*0.0328) ft=0.656 ft

Radius=4/2=2 mm =0.002 m=(0.002*3.28)ft=0.00656 ft

T_1=500 C=932 F

T_2=50 C= 122 F

Part A:

In Joules (J)

$A=\pi *r^2\\A=\pi *(0.002)^2\\A=0.00001256 m^2$

Heat Q is:

$Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{51.88*0.000012566*(500-50}{0.2}\\ Q=1.467 J$

In Btu:

$A=\pi *r^2\\A=\pi *(0.00656)^2\\A=0.00013519 m^2$

Heat Q is:

$Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{8.33*10^{-3}*0.00013519*(932-122}{0.656}\\ Q=0.001390 Btu$

PArt B:

At midpoint Length=L/2=0.1 m

$Q=\frac{k*A*(T_1-T_2)}{L}$

On rearranging:

$T_2=T_1-\frac{Q*L}{KA}$

$T_2=500-\frac{1.467*0.1}{51.88*0.00001256} \\T_2=274.866\ C$

4 0

3 years ago

Timescale limits knowledge for scientists because it is difficult for them to see much beyond their lifetimes. Question 1 option

vaieri [72.5K]

Answer:

I think true

Explanation:

Well I mean...we cant see the future. Certain things will be achieveable in different ganerations like going on mars

8 0

2 years ago

When they say in the United States that a car’s tire is filled “to 32 lb,” they mean that <br>its internal pressure is 32 lbf/i

arsen [322]

Answer:

0.71 lbf

Explanation:

Use ideal gas law:

PV = nRT

where P is absolute pressure,

V is volume,

n is number of moles,

R is universal gas constant,

and T is absolute temperature.

The absolute pressure is the sum of the atmospheric pressure and the gauge pressure.

P = 32 lbf/in² + 14.7 lbf/in²

P = 46.7 lbf/in²

Absolute temperature is in Kelvin or Rankine:

T = 75 + 459.67 R

T = 534.67 R

Given V = 3.0 ft³, and R = 10.731 ft³ psi / R / lb-mol:

PV = nRT

(46.7 lbf/in²) (3.0 ft³) = n (10.731 ft³ psi / R / lb-mol) (534.67 R)

n = 0.02442 lb-mol

The molar mass of air is 29 lbm/lb-mol, so the mass is:

m = (0.02442 lb-mol) (29 lbm/lb-mol)

m = 0.708 lbm

The weight of 1 lbm is lbf.

W = 0.708 lbf

Rounded to two significant figures, the weight of the air is 0.71 lbf.

3 0

3 years ago

A spark ignition engine burns a fuel of calorific value 45MJkg. It compresses the air-ful mixture in accordance with PV^1.3=cons

antoniya [11.8K]

Answer:

i). Compression ratio = 3.678

ii). fuel consumption = 0.4947 kg/hr

Explanation:

Given :

$PV^{1.3}=C$

Fuel calorific value = 45 MJ/kg

We know, engine efficiency is given by,

$\eta = 1-\left ( \frac{1}{r_{c}} \right )^{1.3-1}$

where $r_{c}$ is compression ratio = $\frac{v_{c}+v_{s}}{v_{c}}$

$r_{c}$ = $1+\frac{v_{s}}{v_{c}}$

where $v_{c}$ is compression volume

$v_{s}$ is swept volume

Now it is given that swept volume at 30% of compression, 70% of the swept volume remains.

Then, $v_{30}=v_{c}+0.7v_{s}$

and at 70% compression, 30% of the swept volume remains

∴ $v_{70}=v_{c}+0.3v_{s}$

We know,

$\frac{P_{2}}{P_{1}}=\left ( \frac{V_{1}}{V_{2}} \right )^{n}$

$\frac{2.75}{1.5}=\left ( \frac{v_{c}+0.7\times v_{s}}{v_{c}+0.3\times v_{s}} \right )^{1.3}$

$\left ( 1.833 \right )^{\frac{1}{1.3}}=\frac{v_{c}+0.7v_{s}}{v_{c}+0.3v_{s}}\\$

$1.594=\frac{v_{c}+0.7v_{s}}{v_{c}+0.3v_{s}}$

$v_{c}+0.7v_{s}=1.594v_{c}+0.4782v_{s}$

$0.7v_{s}-0.4782v_{s}=1.594v_{c}-v_{c}$

$0.2218v_{s} = 0.594v_{c}$

$v_{c}=0.3734 v_{s}$

∴ $r_{c}= 1+\frac{v_{s}}{0.3734v_{s}}$

Therefore, compression ratio is $r_{c}$ = 3.678

Now efficiency, $\eta =\left ( 1-\frac{1}{r_{c}} \right )^{0.3}$

$\eta =\left ( 1-\frac{1}{3.678} \right )^{0.3}$

$\eta =0.32342$ , this is the ideal efficiency

Therefore actual efficiency, $\eta_{act} =0.5\times \eta _{ideal}$

$\eta_{act} =0.5\times 0.32342$

$\eta_{act} =0.1617$

Therefore total power required = 1 kW x 3600 J

= 3600 kJ

∴ we know efficiency, $\eta=\frac{W_{net}}{Q_{supply}}$

$Q_{supply}=\frac{W_{net}}{\eta _{act}}$

$Q_{supply}=\frac{3600}{0.1617}$

$Q_{supply}=22261.78 kJ$

Therefore fuel required = $\frac{22261.78}{45000}$

= 0.4947 kg/hr

5 0

3 years ago

1 po

Dmitry_Shevchenko [17]

Answer:

D. Brake

Explanation:

NJMVC is an acronym for New Jersey Motor Vehicle Commission and it is an agency of government that was established in 2003. NJMVC is saddled with the responsibility of inspecting, titling of number plates and registration of motor vehicles, as well as licensing of drivers in New Jersey, United States of America.

The NJMVC subject the applicants (drivers) to a series of test before they are issued a valid driver's license, some of these tests include;

1. Vision test.

2. Knowledge test.

3. Road test.

All of the above mentioned tests must be passed by a driver before he or she are issued a valid driver's license.

Some of the test drive requirements are;

- Valid registration documents.

- Valid sticker of inspection.

- No obstacle or center consoles should prevent the examiner from accessing both the foot brake and parking brake.

Hence, NJMVC will reject a vehicle from the road test if the examiner does not have access to a brake. Thus, the brakes are required to be in good working condition prior to the road test.

However, if an applicant (driver) passes the road test he or she would be issued an authorization for licensing by the examiner and then a digital driver's license by the motor vehicle commission.

5 0

3 years ago