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MatroZZZ [7]
2 years ago
7

Vapor lock occurs when the gasoline is cooled and forms a gel, preventing fuel flow and

Engineering
2 answers:
Andru [333]2 years ago
5 0
The answer is true because vapor locks when the gasoline is cooled and forms a gel preventing fuel flow and engine operation
Sergeu [11.5K]2 years ago
4 0

Answer:

True

Explanation:

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A bus travels the 100 miles between A and B at 50 mi/h and then another 100 miles between B and C at 70 mi/h.
stira [4]

Answer:

c. less than 60 mi/h

Explanation:

To calculate the average speed of the bus, we need to calculate the total distance traveled by the bus, as well as the total time of travel of the bus.

Total Distance Traveled = S = 100 mi + 100 mi

S = 200 mi

Now, for total time, we calculate the times for both speeds from A to b and then B to C, separately and add them.

Total Time = t = Time from A to B + Time from B to C

t = (100 mi)/(50 mi/h) + (100 mi)(70 mi/h)

t = 2 h + 1.43 h

t = 3.43 h

Now, the average speed of bus will be given as:

Average Speed = V = S/t

V = 200 mi/3.43 h

<u>V = 58.33 mi/h</u>

It is clear from this answer that the correct option is:

<u>c. less than 60 mi/h</u>

7 0
3 years ago
What is future active and future passive and future perfect active
san4es73 [151]
The future perfect tense forms are made by putting ‘will / shall + have’ before the past participle from the verb. these sentences can be changed into the passive if the active verb has an object


i hope this helps :D thanks
5 0
3 years ago
Bore = 3"
Grace [21]
I need help my self lol XD
5 0
3 years ago
Read 2 more answers
4.2 A vapor compression refrigeration machine uses 30kW of electric power to produce 50 tons of cooling. What is
stellarik [79]

Answer:

5.833

Explanation:

Coefficient of Perfomance (COP) is the ratio of refrigeration effect to power input.

COP=\frac {RE}{P} where RE is refrigeration effect and P is power input

Here, the power input is given as 30 kW

We also know that 1 ton cooling is equivalent to 3.5 kW hence for 50 tons, RE=50*3.5=175 kW

Now the COP=\frac {175}{30}=5.833

6 0
3 years ago
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
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