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Leona [35]
3 years ago
6

ITS A TESTS PLSSS I NEED THIS RN

Engineering
1 answer:
Travka [436]3 years ago
8 0

Answer:

Explanation:

ok is this multiple choice if so plz provide that and ill answer again

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KIM [24]

Answer:

The solution is given in the attachments.

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3 years ago
What do you enjoy most and least about engineering?
melisa1 [442]

Answer:

“I really love the design work in engineering, the face-to-face interaction with clients, and the opportunity to see projects come to life. But if I had to pick one thing that I don’t enjoy as much, I would have to say it’s contract preparation.”

5 0
3 years ago
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A flow of 100 mgd is to be developed from a 190-mi^2 watershed. At the flow line the area's reservoir is estimated to cover 3900
yaroslaw [1]

Answer:

13-mi 27 acres

Explanation:

7 0
3 years ago
Three 12-V, 100-A-hr batteries are connected in series. What are the output voltage and A-hr capacity of this connection
Lera25 [3.4K]

Answer:

36V, 100A-hr

Explanation:

Since the batteries are connected in series;

i. the output voltage will be the sum of the individual voltages

ii. the current rating (A-hr) will be the same as their individual current rating (A-hr). And this is because the same current flows through the batteries.

From i,

The output voltage, V, is given by the sum of the voltages of the three batteries;

V = 12V + 12V + 12V

V = 36V

From ii,

The A-hr capacity of the connection is the same as that of the individual batteries;

100 A-hr

Therefore, the output voltage and A-hr capacity of this connection is:

36V, 100A-hr

3 0
4 years ago
Air is pumped from a vacuum chamber until the pressure drops to 3 torr. If the air temperature at the end of the pumping process
malfutka [58]

Answer:

The final pressure is 3.16 torr

Solution:

As per the question:

The reduced pressure after drop in it, P' = 3 torr = 3\times 0.133\ kPa

At the end of pumping, temperature of air, T = 5^{\circ}C = 278 K

After the rise in the air temperature, T' = 20^{\circ}C = 293 K

Now, we know the ideal gas eqn:

PV = mRT

So

P = \frac{m}{V}RT

P = \rho_{a}RT          (1)

where

P = Pressure

V = Volume

\rho_{a} = air\ density

R = Rydberg's constant

T = Temperature

Using eqn (1):

P = \rho_{a}RT

\rho_{a} = \frac{P}{RT}

\rho_{a} = \frac{3 times 0.133\times 10^{3}}{0.287\times 278} = 0.005 kg/m^{3}

Now, at constant volume the final pressure, P' is given by:

\frac{P}{T} = \frac{P'}{T'}

P' = \frac{P}{T}\times T'

P' = \frac{3}{278}\times 293 = 3.16 torr

7 0
4 years ago
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