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3 years ago

HELE

Engineering

2 answers:

Neporo4naja [7]3 years ago

6 0

Answer:

D The answer would be D $270

Explanation:

Hope this helps. Please mark me Brainliest

Finger [1]3 years ago

6 0

Answer: A)

Heeeeeeeeeeeeeeeey

Explanation: a. $900

he graph above shows one store's projected demand curve for hand-held vacuum cleaners. If the demand at point P3 is fully

satisfied, how much money will the store make by selling hand-held vacuum cleaners?

hope this helps

brainly plz

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Which of the following is NOT a breach of Netiquette?

alexandr402 [8]

Answer:

I need some more point and i do not understand your question

Explanation:

7 0

3 years ago

What is the average linear (seepage) velocity of water in an aquifer with a hydraulic conductivity of 6.9 x 10-4 m/s and porosit

jeka94

Answer:

a. 0.28

Explanation:

Given that

porosity =30%

hydraulic gradient = 0.0014

hydraulic conductivity = 6.9 x 10⁻4 m/s

We know that average linear velocity given as

$v=\dfrac{K}{n_e}\dfrac{dh}{dl}$

$v=\dfrac{6.9\times 10^{-4}}{0.3}\times0.0014\ m/s$

$v=3.22\times 10^{-6}\ m/s$

The velocity in m/d ( 1 m/s =86400 m/d)

v= 0.27 m/d

So the nearest answer is 'a'.

a. 0.28

4 0

3 years ago

A spring-mass-damper instrument is employed for acceleration measurements. The spring constant is 12000 N/m. The mass is 5 g. Th

shepuryov [24]

Answer:

a) 246.56 Hz

b) 203.313 Hz

c) Add more springs

Explanation:

Spring constant = 12000 N/m

mass = 5g = 5 * 10^-3 kg

damping ratio = 0.4

a) Calculate Natural frequency

Wn = √k/m = $\sqrt{12000 / 5*10^{-3} }$

= 1549.19 rad/s ≈ 246.56 Hz

b) Bandwidth of instrument

W / Wn = $\sqrt{1-2(0.4)^2}$

W / Wn = 0.8246

therefore Bandwidth ( W ) = Wn * 0.8246 = 246.56 * 0.8246 = 203.313 Hz

C ) To increase the bandwidth we have to add more springs

5 0

3 years ago

Question 1 of 8.

Rasek [7]

The answer seems pretty obvious, all of the above

8 0

3 years ago

Read 2 more answers

A gas expands in a piston-cylinder assembly from p1 = 8 bar, V1 = 0.02 m3 to p2 = 2 bar. The relation between pressure and volum

Charra [1.4K]

Answer:

The heat transfer is 29.75 kJ

Explanation:

The process is a polytropic expansion process

General polytropic expansion process is given by PV^n = constant

Comparing PV^n = constant with PV^1.2 = constant

n = 1.2

(V2/V1)^n = P1/P2

(V2/0.02)^1.2 = 8/2

V2/0.02 = 4^(1/1.2)

V2 = 0.02 × 3.2 = 0.064 m^3

W = (P2V2 - P1V1)/1-n

P1 = 8 bar = 8×100 = 800 kPa

P2 = 2 bar = 2×100 = 200 kPa

V1 = 0.02 m^3

V2 = 0.064 m^3

1 - n = 1 - 1.2 = -0.2

W = (200×0.064 - 800×0.02)/-0.2 = -3.2/-0.2 = 16 kJ

∆U = 55 kJ/kg × 0.25 kg = 13.75 kJ

Heat transfer (Q) = ∆U + W = 13.75 + 16 = 29.75 kJ

7 0

3 years ago