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Tresset [83]
3 years ago
13

Which of these is a drawback of hydraulic fracturing?

Physics
2 answers:
11111nata11111 [884]3 years ago
6 0

Answer:

oil and gas drilling across the country. is an example

Explanation:

soldier1979 [14.2K]3 years ago
3 0

Answer:

Oil and Gas would be a drawback

Explanation:

Hope this helps ^^

Have a great day!

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.A coin rolls off the edge of a table. The coin
geniusboy [140]

Answer:

Apply the following formulae horizontally And get A value for time

Remember horizontal acceleration is zero

s  = ut +  \frac{1}{2}a {t}^{2}   \\ 0.8 = 1.7 \times t \\  \frac{0.8}{1.7}  = t \\ t = 0.47s

and then to find the height apply the same above equation vertically...remember vertical initial velocity is zero

s = ut +  \frac{1}{2} a {t}^{2}  \\ s =  \frac{1}{2}  \times 10 \times (0.47) ^{2}  \\ s = 1.1045m

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3 years ago
5. How does privatization contribute to productivity?
tangare [24]

Answer:

increased competition

Explanation:

3 0
2 years ago
Two round rods, one steel andthe other copper, are joined end to end. Each rod is 0.750 mlong and 1.50 cmin diameter. The combin
nikitadnepr [17]

Answer: a) Strain on Steel rod = 0.0001078

b) elongation on the steel rod = 0.00008085m = 0.008085cm = 0.0081mm.

c) strain on Copper Rod = 0.000189

Explanation: a) To obtain the strain of the steel rod, we invoke Hooke's law which states that, provided the elastic limit of A material isn't exceeded, the stress it undergoes is directly proportional to its strain.

(Stress, σ) ∝ (Strain, ε)

The constant of proportionality is called Young's modulus, E.

σ = Eε

For steel, Younger Modulus as obtained from literature = 210GPa.

Strain = Stress/Young's Modulus

Stress = (Force or Load applied)/ Cross sectional Area.

Force applied For the steel = 4000N

Cross sectional Area = (π(D^2))/4

D = 1.50cm = 0.015m

A = 0.0001767 m2

σ = 4000/0.0001767 = 22637238.257 N/m2

ε = σ/E = 22637238.257/(210 × (10^9)) = 0.0001078.

b) To solve for elongation.

Strain, ε = (elongation, dl)/(original length, lo)

Elongation, dl = strain × original length

dl = 0.0001078 × 0.75 = 0.00008085m = 0.008085cm = 0.0081mm.

c) strain in Copper

ε = σ/E; σ = 22637238.257 N/m2

Young's modulus of Copper, from literature, = 120GPa

ε = 22637238.257/(120 × (10^9)) = 0.000189

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3 years ago
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I've asked this question 5 times and still no answer pls help TT
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Answer:

Explanation:

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