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Crazy boy [7]
4 years ago
12

Water is pumped from one large reservoir to another at a higher elevation. If the flow rate is 2.5 ft3 /s and the pump delivers

14 hp, find the energy loss associated with pumping from section 1 to section 2
Engineering
1 answer:
4vir4ik [10]4 years ago
6 0

Answer:

2132hp  ed enregia

e

Explanation:

dawdsawdsawdsawdsawdsawdsawdaas

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An automobile tire with a volume of 0.8 m3 is inflated to a gage pressure of 200 kPa. Calculate the mass of air in the tire if t
saveliy_v [14]

Answer:

2.83 kg

Explanation:

Given:

Volume, V = 0.8 m³

gage pressure, P = 200 kPa

Absolute pressure = gage pressure + Atmospheric pressure

= 200 + 101 = 301 kPa = 301 × 10³ N/m²

Temperature, T = 23° C = 23 + 273 = 296 K

Now,

From the ideal gas equation

PV = mRT

Where,

m is the mass

R is the ideal gas constant = 287 J/Kg K. (for air)

thus,

301 × 10³ × 0.8 = m × 287 × 296

or

m = 2.83 kg

3 0
3 years ago
1. If a bolt is size 1/2" or larger, then its corresponding wrench size should be____ larger than the bolt size
Mrac [35]

Answer:

  1. C. 1/4"
  2. B. 3/16"

Explanation:

1. For hex bolts, lag bolts, and square bolts, the wrench size is 1/4" larger than the bolt size for 1/2" and 9/16" bolts. For 5/8" bolts and larger, the wrench size is <em>50% larger than the bolt size</em>.

__

2. For 7/16" bolts, the wrench size is 5/8", so is 3/16" larger than the bolt. This holds down to 1/4" bolts, where the wrench size may be 3/8" or 7/16".

3 0
4 years ago
A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude
Fofino [41]

Answer:

u = 260.22m/s

S_{max} = 1141.07ft

Explanation:

Given

S_0 = 89.6ft --- Initial altitude

S_{16.5} = 0ft -- Altitude after 16.5 seconds

a = -g = -32.2ft/s^2 --- Acceleration (It is negative because it is an upward movement i.e. against gravity)

Solving (a): Final Speed of the rocket

To do this, we make use of:

S = ut + \frac{1}{2}at^2

The final altitude after 16.5 seconds is represented as:

S_{16.5} = S_0 + ut + \frac{1}{2}at^2

Substitute the following values:

S_0 = 89.6ft       S_{16.5} = 0ft     a = -g = -32.2ft/s^2    and t = 16.5

So, we have:

0 = 89.6 + u * 16.5 - \frac{1}{2} * 32.2 * 16.5^2

0 = 89.6 + u * 16.5 - \frac{1}{2} * 8766.45

0 = 89.6 + 16.5u-  4383.225

Collect Like Terms

16.5u = -89.6 +4383.225

16.5u = 4293.625

Make u the subject

u = \frac{4293.625}{16.5}

u = 260.21969697

u = 260.22m/s

Solving (b): The maximum height attained

First, we calculate the time taken to attain the maximum height.

Using:

v=u  + at

At the maximum height:

v =0 --- The final velocity

u = 260.22m/s

a = -g = -32.2ft/s^2

So, we have:

0 = 260.22 - 32.2t

Collect Like Terms

32.2t = 260.22

Make t the subject

t = \frac{260.22}{ 32.2}

t = 8.08s

The maximum height is then calculated as:

S_{max} = S_0 + ut + \frac{1}{2}at^2

This gives:

S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 32.2 * 8.08^2

S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 2102.22

S_{max} = 89.6 + 260.22 * 8.08 - 1051.11

S_{max} = 1141.0676

S_{max} = 1141.07ft

Hence, the maximum height is 1141.07ft

8 0
3 years ago
When -iron is subjected to an atmosphere of hydrogen gas, the concentration of hydrogen in the iron, CH (in weight percent), is
djyliett [7]

Answer:

See attachments for step by step explanation towards getting answer.

Explanation:

Given that;

College Engineering 10+5 pts

When -iron is subjected to an atmosphere of hydrogen gas, the concentration of hydrogen in the iron, CH (in weight percent), is a function of hydrogen pressure, (in MPa), and absolute pH2temperature (T) according to(5.14)Furthermore, the values of D0 and Qd for this diffusion system are 1.4  10-7 m2/s and 13,400 J/mol, respectively. Consider a thin iron membrane 1 mm thick that is at 250C. Compute the diffusion flux through this membrane if the hydrogen pressure on one side of the membrane is 0.15 MPa (1.48 atm), and on the other side 7.5 MPa (74 atm).

See attachlent for complete solving.

6 0
3 years ago
A zener diode exhibits a constant voltage of 5.6 volts for currents greater than five times the knee current.Izk is specified to
iren2701 [21]

Answer:

<u><em>note:</em></u>

<u><em>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</em></u>

Download docx
5 0
3 years ago
Read 2 more answers
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