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2 years ago

10

The on-board computer controls the suspension system based on inputs from?

1 answer:

lawyer [7]2 years ago

3 0

The answer is A. Sensors

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A scale model is 4th the size of the pump. Determine the power ratio of the pump and its scale model if the ratio of the heads i

vredina [299]

Given:

size of scale model = 4(size of pump)

power ratio of pump and scale model = 5:1

Solution:

Let the diameter of scale model and pump be $d_{s}$ and $d_{p}$ respectively

and head be $H_{s}$ and $H_{p}$ respectively

Now, power, P is given as a function of head(H) and dischagre(Q)

P = $\rho gQH$ (1)

From eqn (1):

$P \propto QH$

and

$QH \propto \sqrt{H}D^{2}$

So,

$P \propto H^{\frac{3}{2}} D^{2}$

Therefore,

$\frac{P_{s}}{P_{p}}$ = $\frac{D_{s}^{2} H_{s}^{\frac{3}{2}}}{D_{p}^{2} H_{p}^{\frac{3}{2}}}$

$\frac{P_{s}}{P_{p}}$ = $\frac{D_{s}^{2} H_{s}^{\frac{3}{2}}}{D_{p}^{2} H_{p}^{\frac{3}{2}}}$

$\frac{P_{s}}{P_{p}}$ = $\frac{1^{2}\times 5^{\frac{3}{2}}}{4^{2}\times 1^{\frac{3}{2}}}$

$\frac{P_{s}}{P_{p}}$ = $\frac{5\sqrt{5}}{16}$

${P_{s}}:{P_{p}}$ = ${5\sqrt{5}}:{16}$

8 0

3 years ago

25°C is flowing in a covered irrigation ditch below ground. Every 100 ft, there is a vent line with a 1.0 in. inside diameter an

Gnesinka [82]

The total evaporation loss of water is 87.873 × $10^{-5}$ lbm /day.

<u>Explanation:</u>

Assume A is the water and B is air.

A is diffusing to non diffusing B.

$N_{A}=\frac{D_{A B} P}{R T Z P_{B/ M}}\left(P_{A_{1}} \ - P_{B_{2}}\right)$

By the table 6.2 - 1 at 25°C, the diffusivity of air and water system is $0.260 \times 10^{-4} \mathrm{m}^{2} / \mathrm{s}$ .

Total pressure P = 1 atm = 101.325 KPa

$P_{A_{1} }$ = 23.76 mm Hg

$P_{A_{1} }$ = $\frac{23.76}{760}$

$P_{A_{1} }$ = 0.03126 atm

$P_{A_{1} }$ = 3.167 K Pa

When air surrounded is dry air, then $PA_{2}$ = 0 mm Hg

R = 8.314 $\frac{K P a \cdot m^{3}}{mol \cdot k}$

$P_{B/M}$ = $\frac{P_{B_{1}}-P_{B_{2}}}{\ln \left(P_{B_{1}} / P_{B_{2}}\right)}$

$P_{B_{1}}=P_{T}-P_{A1_{}}$

= 101.325 - 3.167

$P_{B_{1} }$ = 98.158 K Pa

$P_{B_{2}}=P_{T}-P_{A_{2}}$

$P_{B_{2} }$ = 101.325 - 0

$P_{B_{2} }$ = 101.325 K Pa

$P_{B / M}=\frac{98.158-101.325}{\ln (98.158 / 101 \cdot 325)}$

$P_{B/M}$ = 99.733 K Pa

Z = 1 ft = 0.3048 m

T = 298 K

$N_{A}$ = $\frac{(0.206*10^{-4})(101.325)(3.167 - 0) }{(8.314) ( 298 )( 0.3048 )( 99.733 ) }$

$N_{A}$ = 0.11077 × $10^{-6}$ mol/ $m^{2}$ .s

$N_{A}$ = 0.11077 × $10^{-6}$ × 18 × (60×60×24)

$N_{A}$ = 0.1723 $lb_{m}$ / $m^{2}$ .day

$\tilde{N}_{A}=N_{A} \times Area$

Area of individual pipe is

$A=\frac{\pi}{4}(0.0254)^{2}$

A = 0.00051 $m^{2}$

$\bar{N}_{\boldsymbol{A}}$ = 0.1723 × 0.00051

$\bar{N}_{\boldsymbol{A}}$ = 0.000087873 lbm/day

In 1000 ft length of ditch,there will be a 10 pipes. The amount of evaporation water is

= 10 × 0.000087873 = 0.00087873 lbm /day

The total evaporation loss of water is 87.873 × $10^{-5}$ lbm /day.

5 0

3 years ago

Wheel diameter 150 mm, and infeed 0.06 mm in a surface grinding operation. Wheel speed 1600 m/min, work speed 0.30 m/s, and cros

gogolik [260]

Answer:

a) the average length per chip is 3 mm

b) the metal removal rate MR is 90 mm³/sec

c) number of chips formed per unit time for the portion of the operation when the wheel is engaged in the work is 4,000,000 chips/min

Explanation:

Given that;

wheel diameter = 150 mm

infeed = 0.06 mm

wheel speed = 1600 m/min = 16000000 mm/s

work speed = 0.30 m/s = 300 mm/s

cross feed = 5 mm

active grits per area = 50 grits/cm²

a)

Average length per chip

Average chip length is given by

Lc = √fd

f is infeed and d is diameter of the wheel

so we substitute

Lc = √( 0.06 × 150

Lc = √ 9

Lc = 3 mm

Therefore the average length per chip is 3 mm

b)

metal removal rate MR is expressed as;

MR = fvc

v is work speed, c is cross feed and f is infeed

so we substitute

MR = 0.06 × 300 × 5

MR = 90 mm³/sec

Therefore the metal removal rate MR is 90 mm³/sec

c)

number of chips formed per unit time is expressed as

No = Vw × c × G

Vw is wheel speed and G is active grits per area

so we substitute

No = 1600000 × 5 × 50/10²

= 4,000,000 chips/min

Therefore number of chips formed per unit time for the portion of the operation when the wheel is engaged in the work is 4,000,000 chips/min

4 0

3 years ago

musickatia [10]

Uhh idk this can u mabye look it up

6 0

4 years ago

What would be the most likely scale factor to use for an n-gauge model train setup? (An n-gauge layout uses locomotives that are

slega [8]

A or b for sure brother

8 0

3 years ago