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My name is Ann [436]
2 years ago
11

In surveying , supposing we can not pull the tape because it is passing through a shallow river. What will i do to obtain an acc

urate distance?​
Engineering
1 answer:
natulia [17]2 years ago
7 0

Answer:

Hold the tape in place and go down to the end.

Explanation:

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GrogVix [38]
It would have environmental and societal impacts
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3 years ago
If they opened up the International Space Station to tourism, would you go? Why? answer in 2 sentences
Arlecino [84]
Personally, I would go to the space station. The space station has extreme different levels of technology and abilities, plus who doesn’t want to go to space.
6 0
3 years ago
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Consider casting a concrete driveway 40 feet long, 12 feet wide and 6 in. thick. The proportions of the mix by weight are given
Akimi4 [234]

Answer:

Weight of cement = 10968 lb

Weight of sand = 18105.9 lb

Weight of gravel = 28203.55 lb

Weight of water = 5484 lb

Explanation:

Given:

Entrained air = 7.5%

Length, L = 40 ft

Width,w = 12 ft

thickness,b= 6 inch, convert to ft = 6/12 = 0.5 ft

Specific gravity of sand = 2.60

Specific gravity of gravel = 2.70

The volume will be:

40 * 12 * 0.5 = 240 ft³

We need to find the dry volume of concrete.

Dry volume = wet volume * 1.54 (concrete)

Dry volume will be = 240 * 1.54 = 360ft³

Due to the 7% entarained air content, the required volume will be:

V = 360 * (1 - 0.07)

V = 334.8 ft³

At a ratio of 1:2:3 for cement, sand, and gravel respectively, we have:

Total of ratio = 1+2+3 = 6

Their respective volume will be =

Volume of cement = \frac{1}{6}*334.8 = 55.8 ft^3

Volume of sand = \frac{2}{6}*334.8 = 111.6 ft^3

Volume of gravel = \frac{3}{6}*334.8 = 167.4 ft^3

To find the pounds needed the driveway, we have:

Weight = volume *specific gravity * density of water

Specific gravity of cement = 3.15

Weight of cement =

55.8 * 3.15 * 62.4 = 10968 pounds

Weight of sand =

111.6 * 2.60 * 62.4 = 18105.9 lb

Weight of gravel =

167.4 * 2.7 * 62.4 = 28203.55 lb

Given water to cement ratio of 0.50

Weight of water = 0.5 of weight of cement

= 1/2 * 10968 = 5484 lb

4 0
4 years ago
A gas within a piston–cylinder assembly undergoes an isothermal process at 400 K during which the change in entropy is 20.3 kJ/K
Mashcka [7]

Answer:

W= 8120 KJ

Explanation:

Given that

Process is isothermal ,it means that temperature of the gas will remain constant.

T₁=T₂ = 400 K

The change in the entropy given ΔS = 20.3 KJ/K

Lets take heat transfer is Q ,then entropy change can be written as

\Delta S=\dfrac{Q}{T}

Now by putting the values

20.3=\dfrac{Q}{400}

Q= 20.3 x 400 KJ

Q= 8120 KJ

The heat transfer ,Q= 8120 KJ

From first law of thermodynamics

Q = ΔU + W

ΔU =Change in the internal energy ,W=Work

Q=Heat transfer

For ideal gas ΔU  = m Cv ΔT]

At constant temperature process ,ΔT= 0

That is why ΔU  = 0

Q = ΔU + W

Q = 0+ W

Q=W= 8120 KJ

Work ,W= 8120 KJ

8 0
3 years ago
Mihuv8tr5qwertgyhjzxcvbnfr5y7nnbvcxzwertgyhujio vv solve the riddle
Inessa [10]

Answer:

v1QAZ3EDCRFV5TGB6YHNUJMIK,9OL0K9MIJNUHB7YGVTFCRDXESZWAq

Explanation:

qaAQzwsxedcnujmik,ol mkjuhtfcrxdZSWAQWSEDRFTGYHUJIKO,LP.; ,LMKJNUHTGDXESZWaEDRFTGHJKL,MNBVFDSWQAAWERTYUIOP;L,MNHGFDEWwertyuikolp;[l.,mnbvfre345678990098765434rtyhnbhju8765rtghjui875rfghji8765rfghju7654redfghu7643erfghji987yhjko987y

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3 years ago
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