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Alborosie
2 years ago
15

Technician A says that if fuel pump pressure is correct, fuel pump volume will be correct as well. Technician B says that a fuel

pump may produce specified pressure but below specified volume. Which technician is correct
Engineering
1 answer:
guajiro [1.7K]2 years ago
5 0

Answer:

Technician B only

Explanation:

hope this helps :)

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D. projectile
Colt1911 [192]
Answer: parabola

Explanation:

•Parabolic Trajectory:

In conclusion, projectiles travel with a parabolic trajectory due to the fact that the downward force of gravity accelerates them downward from their otherwise straight-line, gravity-free trajectory.
5 0
3 years ago
Cuáles son los sentidos comunes que se pueden utilizar, para identificar un metal, gases o líquido
Vadim26 [7]

Answer:

A solid has definite volume and shape, a liquid has a definite volume but no definite shape, and a gas has neither a definite volume nor shape.

Explanation:

8 0
3 years ago
An air conditioning performance check requires the use of a
Lostsunrise [7]

Answer:

aA

Explanation:

air conditioning performance check requires the use of aA.

Hope this helps!

4 0
3 years ago
Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a)
Sonbull [250]

Answer:

Explanation:

(a)

\frac{354 mg \, 45 km}{0.0356 kN} = 354 mg \times \frac{1 kg}{10^6 mg} \times 45 km \times \frac{10^3m}{1 km} \times \frac{1}{0.0356 kN} \times \frac{1 kN}{10^3 N} = 0.447 \frac{kg \, m}{N}

(b)

0.00453 Mg \times 201 ms = 0.00453 Mg \times \frac{10^3 kg}{1 Mg} \times 201 ms \times \frac{1 s}{10^3 ms} = 0.911 kg \, s

(c)

\frac{435 MN}{23.2 mm} = 435 MN \times \frac{10^6 N}{1 MN} \times \frac{1}{23.2 mm}  \times \frac{10^3 mm}{1 m} = 18.75 \times 10^9 \frac{N}{m} = 18.75 \frac{GN}{m}

7 0
4 years ago
A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of wid
dmitriy555 [2]

Complete Question:

A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of width 5mm and a depth of cut of 0.5 mm. Estimate the minimum time required to reduce the depth of the plate by 20 mm if the tool moves at 400 mm per second.

Answer:

T_{min} = 26 mins 40 secs

Explanation:

Reduction in depth, Δd = 20 mm

Depth of cut, d_c = 0.5 mm

Number of passes necessary for this reduction, n = \frac{\triangle d}{d_c}

n = 20/0.5

n = 40 passes

Tool width, w = 5 mm

Width of metal plate, W = 200 mm

For a reduction in the depth per pass, tool will travel W/w = 200/5 = 40 times

Speed of tool, v = 100 mm/s

Time/pass = \frac{40*400}{400} \\Time/pass = 40 sec

minimum time required to reduce the depth of the plate by 20 mm:

T_{min} = number of passes * Time/pass

T_{min} = n * Time/pass

T_{min} = 40 * 40

T_{min} =  1600 = 26 mins 40 secs

3 0
4 years ago
Read 2 more answers
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