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3 years ago

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Question 3 Unsaved

2 answers:

natka813 [3]3 years ago

7 0

Sub-surface mining involves digging tunnels and shafts to deep underground.

Answer: Option 4

<u>Explanation: </u>

Mining is the process of extracting ores of minerals from the earth crust depending upon the presence of ores in the earth crust position. Some mineral ores are tend to present in the surface of earth crust.

In order to extract them, surface mining is done. Strip mining and open pit mining are types of surface mining. This type of mining does not require tunnelling and digging deep shafts as the ores can be obtained from the upper layers of earth crust.

Smelting is the process of extracting minerals from the ores, so it is not a process of mining. Thus, sub-surface mining uses digging tunnels and shafts as they are required to extract ores buried in the deep underground of earth crust.

faltersainse [42]3 years ago

5 0

Subsurface mining is the answer it's late though lol

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CaHeK987 [17]

It is A or D but I believe A

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3 years ago

A bug is moving along a meter stick in the negative direction at a constant speed of 0.85cm/s. After 42 seconds have passed the

Agata [3.3K]

Given that,

bug speed, v= 0.85 m/s

time, t =42 s

Final position of bug on meter stick was 27 cm

Starting position of bug on meter stick = ?

Since we know that,

s = vt

s= 0.85*42 = 35.7 cm

this is the distance covered by bug in the given time and velocity.

since the bug is moving in negative direction, starting point will be:

27.0 cm+ 35.7 cm = 62.7 cm

The bugs starting position on meter stick was 62.7 cm.

3 0

4 years ago

Gunther needs a 75% concentrated solution. When making his solution at room temperature, he could only make a solution that was

jolli1 [7]

Answer:

Heat the solution, dissolve the solute, and let it cool verifying nothing settled out.

Explanation:

7 0

3 years ago

Read 2 more answers

Examples of Cardiovascular Endurance ( check all that apply)

sveta [45]

Answer:

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7 0

3 years ago

Assume the average value of the vertical component of Earth's magnetic field is 42 μT (downward) in some region that has an area

Oliga [24]

Answer:

The net magnetic flux through the rest of Earth's surface is $5.782\times10^{-2}\ T$

Explanation:

Given that,

Magnetic field = 42 μT

Area $A=3.71\times10^{5}\ km^2$

$A=3.71\times10^{11}\ m^2$

We need to calculate the flux per unit area

$flux\ per\ unit\ area=\dfrac{42\times10^{-6}}{3.71\times10^{11}}$

$flux\ per\ unit\ area=1.132\times10^{-16}\ T/m^2$

We need to calculate the total earth's surface area

$A'=4\pi r^2$

$A'=4\times\pi\times(6.3781\times10^{6})^2$

$A'=5.1120\times10^{14}\ m^2$

We need to calculate the rest of earth's area

$A''=A-A'$

Put the value into the formula

$A''=5.1120\times10^{14}-3.71\times10^{11}$

$A''=5.10829\times10^{14}\ m^2$

We need to calculate the net magnetic flux through the rest of Earth's surface

$B'=5.10829\times10^{14}\times1.132\times10^{-16}$

$B'=5.782\times10^{-2}\ T$

Hence, The net magnetic flux through the rest of Earth's surface is $5.782\times10^{-2}\ T$

3 0

4 years ago