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Bess [88]
3 years ago
11

What are some homophones​

Engineering
2 answers:
kompoz [17]3 years ago
6 0

Answer:

  1. Hare                Hair
  2. Bear                Bare
  3. Hour               Our
  4. Him                 Hymm
  5. Ail                   Ale
  6. Add                Ad
  7. Fair                Fare
  8. Know             No
  9. Knight            Night
  10. Aisle              Isle

Explanation:

ryzh [129]3 years ago
5 0

Answer:

accessary, accessory.

ad, add.

ail, ale.

air, heir.

aisle, I'll, isle.

all, awl.

allowed, aloud.

alms, arms.

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In a river reach, the rate of inflow at any time is 350 cfs and the rate of outflow is 285 cfs. After 90 min, the inflow and out
Semenov [28]

Answer:

change in storage =  -310,500 ft^3

intital storage= 3.67 acre ft

Explanation:

Given data:

Rate of inflow = 350 cfs

Rate of outflow = 285 cfs

After 90 min,  rate of inflow = 250 cfs

Rate of outflow = 200 cfs

final storage = 10.8 acre-ft

calculating the average inflow and outflow

average inflow  = \frac{(350+250)}{2} = 300 cfs

average outlow  = \frac{(285+200)}{2} = 242.5 cfs

total amount of water drain during the period of one hour

= (average outflow - average inflow) *60*90

= (242.5 - 300)*60*90 = -310,500

change in storage is calculate as

= -310,500 ft^3

in cubic meter

= -310500/35.315 = 8792.30 cm^3

in acre-ft

= -310,500/43560 = 7.13 acre ft

initial storage = 10.8 - 7.13 = 3.67 acre ft

3 0
3 years ago
PLEASE HELP
lyudmila [28]

Answer:

Air

Explanation:

6 0
3 years ago
Read 2 more answers
1. A soil core sampling tube of 4 cm diameter, 12 cm length and initial mass of 0.525 kg (sample only), was dried at 105o C and
belka [17]

Answer:

porosity = 0.07 or 7%

dry bulk density = 3.25g/cm3]

water content =

Explanation:

bulk density = dry Mass / volume of  sample

dry mass = 0.490kg = 490g

volume = πr2h = 3.142 * 2 *2 *12 = 150.8cm3

density = 490/150.8 = 3.25g/cm3

porosity = \frac{wet mass - dry mass }{wet mass} = \frac{0.525 - 0.49}{0.525} = 0.07 or 7%

water content =  \frac{wet mass - dry mass}{wet mass} = 7%

8 0
4 years ago
Read 2 more answers
I am trying to create a line of code to calculate distance between two points. (distance=[tex]\sqrt{ (x2-x1)^2+(y2-y1)^2}) My li
k0ka [10]

Answer:

point_dist = math.sqrt((math.pow(x2 - x1, 2) + math.pow(y2 - y1, 2))

Explanation:

The distance formula is the difference of the x coordinates squared, plus the difference of the y coordinates squared, all square rooted.  For the general case, it appears you simply need to change how you have written the code.

point_dist = math.sqrt((math.pow(x2 - x1, 2) + math.pow(y2 - y1, 2))

Note, by moving the 2 inside of the pow function, you have provided the second argument that it is requesting.

You were close with your initial attempt, you just had a parenthesis after x1 and y1 when you should not have.

Cheers.

6 0
3 years ago
Why is Ethics an important part of engineering?
NemiM [27]

Answer:

Products created can change society, for better and for the worse

It was damaged in a fire

8 0
3 years ago
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