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FrozenT [24]
2 years ago
15

HELP!!

Chemistry
1 answer:
irakobra [83]2 years ago
8 0

Answer: In physics, potential energy is the energy held by an object because of its position relative to other objects, stresses within itself, its electric charge, or other factors.

Explanation:

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Consider the relationship (y+3)2 = b/(x-2), where y and x are variables and bis a constant. On rectangular coordinate paper, wha
Nikolay [14]

Answer:

(-1) is the slope of a graph of In(y+3) on the vertical axis versus In(x-2) on the horizontal axis.

Explanation:

\frac{(y+3)}{2} = \frac{b}{(x-2)}

Taking natural logarithm on both the sides:

\ln [(y+3)]-\ln[2]=\ln [b]-\ln [(x-2)]

\ln [(y+3)]=\ln[2]+\ln [b]-\ln [(x-2)]

\ln [(y+3)]=\ln {[2\times b]-\ln [(x-2)]

Slope intercept form is generally given as:

y=mx+c

m = slope, c  = intercept on y axis or vertical axis

On rearranging equation:

\ln [(y+3)]=(-1)\times \ln [(x-2)]+\ln {2b}

y = ln [(y+3)], x = ln [(x-2)], m=-1 , c  = ln 2b

(-1) is the slope of a graph of In(y+3) on the vertical axis versus In(x-2) on the horizontal axis.

8 0
3 years ago
Which best describes how magnesium (Mg) and oxygen (O) bond?
Ksivusya [100]

Answer:

a

They form an ionic bond by exchanging two electrons.

Explanation:

Magnesium loses two electrons and oxygen gains these electrons so ionic bond is formed

8 0
2 years ago
A horizontal pipe 15 cm in diameter and 4 m long is buried in the earth at a depth of 20 cm. The pipe-wall temperature is 75 deg
Basile [38]

Explanation:

The given data for case (1) is as follows.

  h = 20 cm = 0.2 m

Assuming that a rectangular slab is placed above the pipe and we will calculate the heat transfer as follows.

              Q = kA \frac{\Delta T}{L}

  where,    A = area

                  L = length

                  k = thermal conductivity = 0.8 W/m

             \Delta T = change in temperature.

Therefore, putting the given values into the above formula as follows.

            Q = kA \frac{\Delta T}{L}

                = 0.8 W/m \times (4 m \times 0.15 m) \frac{75 - 5}{0.2}

                = 168 W

For case (2), h = 180 cm = 1.8 m

Therefore, heat lost will be calculated as follows.

                       Q = kA \frac{\Delta T}{L}

                           = 0.8 W/m \times (4 m \times 0.15 m) \frac{75 - 5}{1.8}

                           = 18.67 W

Thus, we can conclude that 18.67 W heat lost if the pipe was buried at a depth of 180 cm.

8 0
3 years ago
Q7-A graduated cylinder is filled to the 12.0 mL line with water. A solid with a mass of 14.52 g is placed in the graduated cyli
Tems11 [23]

Answer:

If it served you, give me 5 stars please, thank you!

<h3><u>c) 13.29 mL</u></h3>

6 0
2 years ago
What is the expected mass (in kg) of a 10 over 5 B isotope? 1 proton = 1.6726 × 10-27 kg 1 neutron = 1.6749 × 10-27 kg A.1.67 ×
Ira Lisetskai [31]
^{10}_5 B
An atom of this isotope contains 5 protons and 10-5=5 neutrons.

5 \times 1.6726 \times 10^{-27} + 5 \times 1.6749 \times 10^{-27} = \\&#10;8.363 \times 10^{-27} + 8.3745 \times 10^{-27}= \\&#10;16.7375 \times 10^{-27}= \\&#10;1.67375 \times 10^{-26} \approx \\&#10;1.67 \times 10^{-26}

The answer is A. 1.67 × 10⁻²⁶ kg.
8 0
3 years ago
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