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3 years ago

13

Which of the following is NOT a problem associated with surface mining?

1 answer:

just olya [345]3 years ago

6 0

<h2>Question:</h2>

Which of the following is NOT a problem associated with surface mining?

<h2>Answer:</h2>

<u>Oil</u><u> </u><u>spills</u><u> </u>

<h2><u>#CARRYONLEARNING</u><u> </u></h2><h2><u>#STUDYWELL</u><u> </u></h2>

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The principal value of a Pareto diagram is as a

vlada-n [284]

The Pareto principle is that most things in our life are not commonly distributed.

<u>Explanation:</u>

Pareto chart shows that most of the things which we have in our life and the resources in our life are not equally distributed. The ratio is not always 50:50 according to this principle.

The most important use of a Pareto diagram is to show the most important factor among the set of factors that have been shown. Along with that it also shows the sources which lead to the common defects in the system and tries to solve those defects which occur most often.

4 0

4 years ago

A thick aluminum block initially at 26.5°C is subjected to constant heat flux of 4000 W/m2 by an electric resistance heater whos

Yanka [14]

Given Information:

Initial temperature of aluminum block = 26.5°C

Heat flux = 4000 w/m²

Time = 2112 seconds

Time = 30 minutes = 30*60 = 1800 seconds

Required Information:

Rise in surface temperature = ?

Answer:

Rise in surface temperature = 8.6 °C after 2112 seconds

Rise in surface temperature = 8 °C after 30 minutes

Explanation:

The surface temperature of the aluminum block is given by

$T_{surface} = T_{initial} + \frac{q}{k} \sqrt{\frac{4\alpha t}{\pi} }$

Where q is the heat flux supplied to aluminum block, k is the conductivity of pure aluminum and α is the diffusivity of pure aluminum.

After t = 2112 sec:

$T_{surface} = 26.5 + \frac{4000}{237} \sqrt{\frac{4(9.71\times 10^{-5}) (2112)}{\pi} }\\\\T_{surface} = 26.5 + \frac{4000}{237} (0.51098)\\\\T_{surface} = 26.5 + 8.6\\\\T_{surface} = 35.1\\\\$

The rise in the surface temperature is

Rise = 35.1 - 26.5 = 8.6 °C

Therefore, the surface temperature of the block will rise by 8.6 °C after 2112 seconds.

After t = 30 mins:

$T_{surface} = 26.5 + \frac{4000}{237} \sqrt{\frac{4(9.71\times 10^{-5}) (1800)}{\pi} }\\\\T_{surface} = 26.5 + \frac{4000}{237} (0.4717)\\\\T_{surface} = 26.5 + 7.96\\\\T_{surface} = 34.5\\\\$

The rise in the surface temperature is

Rise = 34.5 - 26.5 = 8 °C

Therefore, the surface temperature of the block will rise by 8 °C after 30 minutes.

5 0

3 years ago

Martha has been running a small business for two years. She now seeks additional investment to finance her business. She has fou

Dafna11 [192]

Answer:

The correct option is B) Balance Sheet

Explanation:

A Balance Sheet offers a description of a company's obligations, assets, and investments as well as net income over a given span of time such as a period of 6 months or 12 months, for instance.

Also known as the Statement of Financial Position, it contains sufficient information for investors and business owners to determine the company's financial performance in that period as well as to compare the performance of that company with industry norms or competition.

Cheers

8 0

3 years ago

A center-point bending test was performed on a 2 in. x d in. wood lumber according to ASTM D198 procedure with a span of 4 ft an

Zigmanuir [339]

Answer:

3.03 INCHES

Explanation:

According to ASTM D198 ;

Modulus of rupture = ( M / I ) * y ----- ( 1 )

M ( bending moment ) = R * length of span / 2

= (120 * 10^3 ) * 48 / 2 = 288 * 10^4 Ib-in

I ( moment of inertia ) = bd^3 / 12

= ( 2 )*( d )^3 / 12 = 2d^3 / 12

b = 2 in , d = ?

length of span = 4 * 12 = 48 inches

R = P / 2 = 240 * 10^3 / 2 = 120 * 10^3 Ib

y ( centroid distance ) = d / 2 inches

back to equation ( 1 )

( M / I ) * y

940.3 ksi = ( 288 * 10^4 / 2d^3 / 12 ) * d / 2

= ( 288 * 10^4 * 12 ) / 2d^3 ) * d / 2

940300 = 34560000* d / 4d^3

4d^3 ( 940300 ) = 34560000 d ( divide both sides with d )

4d^2 = 34560000 / 940300

d^2 = 9.188 ∴ Value of d ≈ 3.03 in

8 0

3 years ago

An aluminum alloy tube with an outside diameter of 3.50 in. and a wall thickness of 0.30 in. is used as a 14 ft long column. Ass

slega [8]

Answer:

slenderness ratio = 147.8

buckling load = 13.62 kips

Explanation:

Given data:

outside diameter is 3.50 inc

wall thickness 0.30 inc

length of column is 14 ft

E = 10,000 ksi

moment of inertia $= \frac{\pi}{64 (D_O^2 -D_i^2)}$

$I = \frac{\pi}{64}(3.5^2 -2.9^2) = 3.894 in^4$

Area $= \frac{\pi}{4} (3.5^2 -2.9^2) = 3.015 in^2$

$radius = \sqrt{\frac{I}{A}}$

$r = \sqrt{\frac{3.894}{3.015}$

r = 1.136 in

slenderness ratio $= \frac{L}{r}$

$= \frac{14 *12}{1.136} = 147.8$

buckling load $= P_cr = \frac{\pi^2 EI}}{l^2}$

$P_{cr} = \frac{\pi^2 *10,000*3.844}{( 14\times 12)^2}$

$P_{cr} = 13.62 kips$

3 0

3 years ago