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3 years ago

13

Transmission cleaners are used: A) Only in conjunction with fuel system cleanersB) Only in the colder monthsC) By themselvesD) I

n conjunction with a transmission fluid exchange

2 answers:

valentina_108 [34]3 years ago

5 0

Answer: D

Explanation: When you change your transmission fluid you change your filter. When you do that you clean where the filter was. Then you can put the new filter on and the new fluid.

irinina [24]3 years ago

3 0

Explanation: When you change your transmission fluid you change your filter. When you do that you clean where the filter was. Then you can put the new filter on and the new fluid.

is correct

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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]

<u>Explanation:</u>

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

For a fluid with a Prandtl Number of 1000.0, the hydrodynamic layer is thinner than the thermal boundary layers. a) True b) Fals

kvv77 [185]

Answer:

(b)False

Explanation:

Given:

Prandtl number(Pr) =1000.

We know that $Pr=\dfrac{\nu }{\alpha }$

Where $\nu$ is the molecular diffusivity of momentum

$\alpha$ is the molecular diffusivity of heat.

Prandtl number(Pr) can also be defined as

$Pr=\left (\dfrac{\delta }{\delta _t}\right )^3$

Where $\delta$ is the hydrodynamic boundary layer thickness and $\delta_t$ is the thermal boundary layer thickness.

So if Pr>1 then hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.

In given question Pr>1 so hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.

So hydrodynamic layer will be thicker than the thermal boundary layer.

8 0

3 years ago

A standard penetration test has been conducted on a coarse sand at a depth of 16 ft below the ground surface. The blow counts ob

scoray [572]

Solution :

Given :

The number of blows is given as :

0 - 6 inch = 4 blows

6 - 12 inch = 6 blows

12 - 18 inch = 6 blows

The vertical effective stress $=1500 \ lb/ft^2$

$= 71.82 \ kN/m^2$

$\sim 72 \ kN/m^2$

Now,

$N_1=N_0 \left(\frac{350}{\bar{\sigma}+70} \right)$

$N_1 =$ corrected N - value of overburden

$\bar{\sigma}=$ effective stress at level of test

0 - 6 inch, $N_1=4 \left(\frac{350}{72+70} \right)$

= 9.86

6 - 12 inch, $N_1=6 \left(\frac{350}{72+70} \right)$

= 14.8

12 - 18 inch, $N_1=6 \left(\frac{350}{72+70} \right)$

= 14.8

$N_{avg}=\frac{9.86+14.8+14.8}{3}$

= 13.14

= 13

8 0

3 years ago

Information such as tolerances and scale can be found in the _______________ of an engineering drawing.

nasty-shy [4]

Answer:

Information such as tolerance and scale can be found in the <u>title block</u> of an engineering drawing

Explanation:

The title block of an engineering drawing can normally be found on the lower right and corner of an engineering drawing and it carries the information that are used to specify details that are specific the drawing including, the name of the project, the name of the designer, the name of the client, the sheet number, the drawing tolerance, the scale, the issue date, and other relevant information, required to link the drawing with the actual structure or item

8 0

3 years ago

Consider the diffusion of water vapor through a polypropylene (PP) sheet 1 mm thick. The pressures of H2O at the two faces are 3

Neko [114]

Answer:

$\boxed{0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}}$

Explanation:

Diffusion flux of a gas, J is given by

$J=P_m\frac {\triangle P}{\triangle x}$ where $P_m$ is permeability coefficient, $\triangle$ P is pressure difference and x is thickness of membrane.

The pressure difference will be 10,000 Pa- 3000 Pa= 7000 Pa

At 298 K, the permeability coefficient of water vapour through polypropylene sheet is $38\times 10^{-13}(cm^{3}. STP)(cm)/(cm^{2}.s.Pa)$

Since the thickness of sheet is given as 1mm= 0.1 cm then

$J=38\times 10^{-13}(cm^{3}. STP)(cm)/(cm^{2}.s.Pa)\times \frac {7000 pa}{0.1cm}=0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}$

Therefore, the diffusion flux is $\boxed{0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}}$

7 0

3 years ago