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3 years ago

Please help meeeee !

Engineering

2 answers:

harina [27]3 years ago

8 0

Do u still need help?

Lena [83]3 years ago

4 0

Oof, that looks complicated. Btw it's been a week do you still need the answer?

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Two Technicians are discussing ShopKey Pro. Technician

Leno4ka [110]

Answer:

Technician B

Explanation:

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3 years ago

Joe Bruin has a big lawn in front of his house that is 30 meters wide and 20 meters long. Josephine makes him go out and mow the

zysi [14]

Explanation:

5 Horsepower for 30 mins,

(5)(745.7) = 3.7285 KW power delivered

General Efficiency of IC engine = 20%

Power required = $\frac{3 \cdot 7285}{0 \cdot 2}=18 \cdot 6425 kw$

Energy required per week,

=P × Time = 18.64 × 60 × 30 = 33.5565 MJ

Lawn area = (30) (20) = 600 $m^{2}$

let sunlight hours be 8 hours

Hence, solar power input on lawn,

=5.62×3600 = 20232 kJ/ $m^{2}$ /day

energy input in lawn = (600) (20232) (7)

= 84974.4 mJ/week

Chemical efficiency by photosynthesis = 4%

Chemical content in grass = (84974.4) (0.04)

= 3398.97 mJ

Mass of the clippers $=(30)(20)(1 \cdot 096)^{2}(667)$

$=478632 \cdot 33$ pounds

Removing water content,

dried grass clippings $$=95726.46$ pound$

= 11533.25 gallons

Trash cans repaired

$=\frac{11533}{50} =230.66\\=231 cans$

By burning the gas, total energy input = 3398.97 MJ × 0.2

= 679.794 MJ

Efficiency of steeling engine = 20%

Energy output by engine = 679.794 ×0.2

= 135.96 mJ

Energy required by mover = 33.5565 mJ

Hence, Energy (output) ⇒ energy required

5 0

3 years ago

What engine or car do you think gets better gas mileage a

defon

Answer:

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Explanation:

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3 years ago

On-site oil storage containers must be marked "Used Oil." True. False.

Scrat [10]

Answer:

How must used oil storage containers be marked? Containers and aboveground tanks used to store used oil at generator facilities must be labeled or marked clearly with the words “Used Oil" (40 CFR Section 279.22(c)).

Explanation:

i think it will help you

7 0

3 years ago

Velocity components in an incompressible flow are: v = 3xy + x^2 y: w = 0. Determine the velocity component in the x-direction.

cupoosta [38]

Answer:

Velocity component in x-direction $u=-\frac{3}{2}x^2-\frac{1}{3}x^3$ .

Explanation:

v=3xy+ $x^{2}$ y

We know that for incompressible flow

$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$

$\frac{\partial v}{\partial y}=3x+x^{2}$

So $\frac{\partial u}{\partial x}+3x+x^{2}=0$

$\frac{\partial u}{\partial x}= -3x-x^{2}$

By integrate with respect to x,we will find

$u=-\frac{3}{2}x^2-\frac{1}{3}x^3$ +C

So the velocity component in x-direction $u=-\frac{3}{2}x^2-\frac{1}{3}x^3$ .

3 0

3 years ago