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2 years ago

7

Situation: Peter is designing a new hybrid car that functions on solar power. He is currently working on sketches of his design

using geometric patterns and shapes.

1 answer:

givi [52]2 years ago

8 0

Answer:

whats the question?

Explanation:

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The high-pressure fuel pumps used in gasoline direct-injection (GDI) systems are powered by ________.

lidiya [134]

Answer:

Camshaft

Explanation:

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3 years ago

A slight breeze is blowing over the hot tub above and yields a heat transfer coefficient h of 20 W/m2 -K. The air temperature is

patriot [66]

Answer:4050 W

Explanation:

Given

Heat transfer Coefficient(h)= $20 W/m^2-K$

Air temperature =75 F

surface area(A)= $7.5 m^2$

Temperature of hot tube is 102 F

We know heat transfer due to convection is given by

$Q=hA\left ( \Delta T\right )$

$Q=20\times 7.5\left ( 102-75\right )=4050 W$

7 0

3 years ago

What are the standard procedures involved in the fixing/securing of cables?

Sladkaya [172]

<u>Cable should be pre-cut and hung suspended for 48 hours to develop its most natural set and lay prior to installation.</u>

<u>Cable should be installed with, not against, its natural set. ... </u>

<u>Strain relief on either end will reduce conductor breakage at the flex points.</u>

4 0

2 years ago

Reverse masking forms a soft edge on the panel.

valina [46]

True I think I’m not sure?

7 0

2 years ago

A certain part of the cast iron piping of a water distribution system involves a parallel section. Both parallel pipes have a di

Bezzdna [24]

Answer :

<h3>Flow rate in pipe B is = 0.3094 $\frac{m^{3} }{s}$ </h3>

Explanation:

Given :

Length of pipe A $L_{A} = 1500$ m

Length of pipe B $L_{B} = 2500$ m

Flow rate through pipe A $Q_{A} = 0.4 \frac{m^{3} }{s}$

Diameter of pipe $D = 30 \times 10^{-2}$ m

Velocity from pipe A,

$V _{A} = \frac{Q_{A} }{A}$

$V _{A} = \frac{0.4 \times 4 }{\pi ( 30 \times 10^{-2} )^{2} }$

$V_{A} = 5.66$ $\frac{m}{s}$

Here, head loss is same because height is same.

$h_{a} = h_{b}$

$L_{A} V_{A} ^{2} = L_{B} V_{B} ^{2}$

$V_{B} = \sqrt{\frac{1500}{2500}} (5.66)$

$V_{B} = 4.38$ $\frac{m}{s}$

Now rate of flow from pipe B is,

$Q_{B} = V_{B} A$

$Q_{B} = \frac{\pi }{4} (0.3)^{2} \times 4.38$

$Q_{B} = 0.3094$ $\frac{m^{3} }{s}$

4 0

3 years ago