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kobusy [5.1K]
2 years ago
6

Which type of mine involves digging tunnels and shafts deep underground?

Physics
1 answer:
mel-nik [20]2 years ago
7 0

The mine involves digging tunnels and shafts deep underground is the underground mining.

<h3>What is mining?</h3>

Mining is the digging of the Earth deep down in search of some precious elements or resources found on Earth.

Underground mining is the digging down into the earth in order to create tunnels. The shafts need to be inserted into those tunnels so that it reaches the deposits of resources.

Thus, underground mining involves digging tunnels and shafts deep underground.

Learn more about mining.

brainly.com/question/14277327

#SPJ4

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What voltage would be measured across the 15 ohm resistor?
algol13

Answer:

7.5 volts

Explanation:

I did it on USA Testprep

5 0
3 years ago
PLEASE HELP!! WILL REWARD!!
Fudgin [204]

velocity = \frac{displacement}{time}

Displacement measures how far it is from the starting point and the final location, which in this case is the distance between Honolulu and Los Angeles. The difference between distance and displacement is that distance measures how far the object actually travels. For example, the airplane may not fly in a straight line and it may have detours, making the distance much greater than the displacement.

On the other hand, the plane could fly at different speeds at different periods of time. It will be so much more complicated if we have to calculate it. That's why this question only asks you to calculate the average velocity of the plane, ie. \frac{total displacement}{total time}

So, we have

the required average velocity = \frac{2558}{5}

= 511.6 miles/hour

Hope this helps!

6 0
3 years ago
Read 2 more answers
The engine starter and a headlight of a car are connected in parallel to the 12.0-V car battery. In this situation, the headligh
stepladder [879]

Answer:

The total power they will consume in series is approximately 2.257 W

Explanation:

The connection arrangement of the headlight and the engine starter = Parallel to the battery

The voltage of the battery, V = 12.0 V

The power at which the headlight operates in parallel, P_{headlight} = 38 W

The power at which the kick starter operates in parallel, P_{kick \ starter} = 2.40 kW

We have;

P = V²/R

Where;

R = The resistance

V = The voltage = 12 V (The voltage is the same in parallel circuit)

For the headlight, we have;

R₁ = V²/P_{headlight}  = 12²/38 = 72/19

R₁ = 72/19 Ω

For the kick starter, we have;

R₂ = V²/P_{kick \ starter} = 12²/2.4 = 60

R₂ = 60 Ω

When the headlight and kick starter are rewired to be in series, we have;

Total resistance, R = R₁ + R₂

Therefore;

R = ((72/19) + 60) Ω = (1212/19) Ω

The current flowing, I = V/R

∴ I = 12 V/(1212/19) Ω = (19/101) A

We note that power, P = I²R

In the series connection, we have;

P_{headlight} = I² × R₁

∴ P_{headlight} = ((19/101) A)² × 72/19 Ω = 1368/10201 W ≈ 0.134 W

The power at which the headlight operates in series, P_{headlight, S} ≈ 0.134 W

P_{kick \ starter} = ((19/101) A)² × 60 Ω = 21660/10201 W ≈ 2.123 W

The power at which the kick starter operates in series, P_{kick \ starter, S} ≈ 2.123 W

The total power they will consume, P_{Total} = P_{headlight, S} + P_{kick \ starter, S}

Therefore;

P_{Total} ≈ 0.134 W + 2.123 W = 2.257 W

4 0
3 years ago
If it has enough kinetic energy, a molecule at the surface of the Earth can "escape the Earth's gravitation", in the sense that
user100 [1]

Answer:

K_E=mgr_E

Explanation:

By conservation of energy, the sum of the kinetic and gravitational potential energies at the surface of the Earth must be equal than their sum at infinity, so we have:

K_E+U_E=K_\infty+U_\infty

\frac{mv_E^2}{2}-\frac{GM_Em}{r_E}=\frac{mv_\infty^2}{2}-\frac{GM_Em}{r_\infty}

Where G=6.67\times10^{-11}Nm^2/kg^2 is the gravitational constant,M_E=5.97\times10^{24}kg and r_E=6371000m are the mass and radius of the Earth, <em>m </em>is the mass of the particle, v_E its velocity at the surface of the Earth (which would be its escape velocity) and v_\infty and r_\infty are the velocities and distance at infinity, which would be null and infinity respectively, so the right hand side of our equation is 0J, which leaves us with:

\frac{GM_Em}{r_E}=\frac{mv_E^2}{2}=K_E

Also, since the force the molecule experiments is the force of gravity (disregarding drag), we can write its weight in terms of Newton's Law of Gravitation:

F=mg=\frac{GM_Em}{r_E^2}

Which means that:

\frac{GM_Em}{r_E}=mgr_E

So finally putting all together we can write:

K_E=\frac{GM_Em}{r_E}=mgr_E

4 0
3 years ago
A narrow wire is used to slow the flow of the electrons in a circuit
vladimir2022 [97]
That's true.

In fact, the resistance of a wire is given by:
R= \frac{\rho L}{A}
where
\rho is the resistivity of the material
L is the length of the wire
A is the cross-sectional area of the wire

We see that the resistance of the wire is inversely proportional to the cross-sectional area: A. Therefore, the narrower the wire, the smaller A, the larger the resistance. But higher resistance means that the current flowing through the wire is lower, therefore the flow of electrons in the circuit is slower, and the initial sentence is true.
3 0
3 years ago
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