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sp2606 [1]
2 years ago
10

Zoe read that drinking water from a faucet instead of from a plastic bottle helps overcome a disadvantage of using nonrenewable

resources. Which conclusion is best supported by this information?
Water can be used up and cannot be replaced.
Plastic is made from petroleum oil.
Making plastics is inexpensive.
The water in faucets in healthier.
Physics
1 answer:
Arlecino [84]2 years ago
7 0

The conclusion that is best supported by this information is the water in faucets in healthier.

<h3>Which conclusion is best supported by this information?</h3>

The water in faucets in healthier is the conclusion that is best supported by this information because plastic made of chemical so water can react with the water and make water unhealthier for us.

So we can conclude that the water in faucets in healthier is the conclusion that is best supported by this information

Learn more about water here: brainly.com/question/1313076

#SPJ1

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Felix has a bucket of golf balls. The table shows the number of golf balls of each color in the
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Answer:c

Explanation:

4 0
3 years ago
A Cessna 150 aircraft has a lift-off speed of approximately 125 km/h. What minimum constant acceleration does this require if th
aivan3 [116]

Answer:

Part a)

a = 3.65 m/s^2

Part b)

t = 9.5 s

Part c)

v_f = 55.1 m/s

Explanation:

Part a)

As we know that it starts from rest and moves on runway by total distance 165 m

so we will have

v_f^2 - v_i^2 = 2ad

v_f^2 - 0 = 2(a)(165)

v_f = 125 km/h = 34.7 m/s

now we have

a = 3.65 m/s^2

Part b)

Now for take off time we will have

v_f - v_i = at

34.7 - 0 = 3.65 t

t = 9.5 s

Part c)

v_f = v_i + at

v_f = 0 + (3.65)(15.1)

v_f = 55.1 m/s

7 0
3 years ago
Whu is it important to mave regular supervision of weights and measurements in the market​
pentagon [3]

Answer:

Hope the answer helped you. if yes pls follow me

Explanation:

the fundamental answer is without regular supervision of definition of weights and measures,commerce exchange will be impossible and there would be no market whatsoever for anything.

5 0
3 years ago
a closed systems internal energy changes by 178 j as a result of being heated with 658 j of energy. the energy used to do work b
Akimi4 [234]
<h3><u>Answer;</u></h3>

= 480 Joules

<h3><u>Explanation;</u></h3>

We use the formula, Q - W = ΔU

Where, Q = Heat transferred to the system

W = Work done by the system

ΔU = Change of internal energy.

As per the question, Q = 658 J

ΔU = 178 J

Thus, W = Q - ΔU = (658 - 178) J = 480 J.

The energy used to do work by the system is 480 J.

7 0
3 years ago
During a compaction test in the lab a cylindrical mold with a diameter of 4in and a height of 4.58in was filled. The compacted s
Ray Of Light [21]

Answer:

part a : <em>The dry unit weight is 0.0616  </em>lb/in^3<em />

part b : <em>The void ratio is 0.77</em>

part c :  <em>Degree of Saturation is 0.43</em>

part d : <em>Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb</em>

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}

Here

  • \gamma_d is the dry unit weight which is to be calculated
  • γ is the bulk unit weight given as

                                              \gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3

  • w is the moisture content in percentage, given as 12%

Substituting values

                                              \gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3

<em>The dry unit weight is 0.0616  </em>lb/in^3<em />

Part b

Void Ratio

The void ratio is given as

                                                e=\frac{G_s \gamma_w}{\gamma_d} -1

Here

  • e is the void ratio which is to be calculated
  • \gamma_d is the dry unit weight which is calculated in part a
  • \gamma_w is the water unit weight which is 62.4 lb/ft^3 or 0.04 lb/in^3
  • G is the specific gravity which is given as 2.72

Substituting values

                                              e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766

<em>The void ratio is 0.77</em>

Part c

Degree of Saturation

Degree of Saturation is given as

S=\frac{G w}{e}

Here

  • e is the void ratio which is calculated in part b
  • G is the specific gravity which is given as 2.72
  • w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

                                      S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261

<em>Degree of Saturation is 0.43</em>

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}

Here

  • \gamma_{zav} is  the zero air unit weight which is to be calculated
  • \gamma_w is the water unit weight which is 62.4 lb/ft^3 or 0.04 lb/in^3
  • G is the specific gravity which is given as 2.72
  • w is the moisture content in percentage, given as 12% or 0.12 in fraction

                                      \gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}\\\gamma_{zav}=\frac{0.04}{0.12+\frac{1}{2.72}}\\\gamma_{zav}=\frac{0.04}{0.4876}\\\gamma_{zav}=0.08202 lb/in^3\\

Now as the volume is known, the the overall weight is given as

weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.

4 0
3 years ago
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