Answer:
Thin, aluminium and buried underground.
Explanation:
When it comes to electrification of a state or province, some characteristics of the wire to use must be considered. This would help to minimize and avoid power loss and wire burns.
i. The wire to use should be thin, and a quite number can be twisted one against the other so as to increase the surface area for heat dissipation.
ii. Aluminium wire is more preferable for this project. It has a high melting point, and reduces energy loss.
iii. Burying the wire underground through an insulator is the best choice, though expensive but would preserve the wire from external influence.
Before comparing and contrasting these layers of Earth, we first define what lithosphere and asthenosphere are.
Lithosphere primarily consists of the outermost layers of the Earth, which are the crust and the uppermost portion of the mantle. Simply, the ground you stepped on is part of earth's lithosphere. On the other hand, asthenosphere comprises of hot and partially molten rock just located at the upper portion of the mantle but just below the lithosphere. Both have similarities and differences, which are as follows:
SIMILARITIES:
- Both are the passageways of earthquakes P-waves (Primary waves) just before it reaches the earth's surface.
- Both are made of the same material (Silicon oxide rocks, which are rich in iron and magnesium)
DIFFERENCES:
- Rocks in lithosphere can bend (it deforms, resulting in fault formations), however, rocks in the asthenosphere, not only bend but also flow (plastic in nature).
- Lithosphere has relatively low temperatures compared to asthenosphere.
- Due to its depth, pressure against rocks in asthenosphere is comparatively higher compared to lithosphere.
The volume of the gas is 
Explanation:
At standard temperature and pressure (stp), the volume occupied by 1 mole of an ideal gas is always equal to

In this problem, we have an amount of moles of gas of
n = 0.034 mol
Since 1 mol of gas occupies a volume of 22.4 L, we can set up the following ratio:

where V is the volume occupied by 0.034 mol of gas. Solving for V, we find

Or, in cubic metres,

Learn more about ideal gases:
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Given: Normal pull of gravity g = 9.8 m/s²;
g = 0.855 m/s² (at a certain distance)
Universal gravitational constant G = 6.67 x 10⁻¹¹ N.m²/Kg²
Mass of the Earth Me = 5.98 x 10²⁴ Kg
Radius r = ?
g = GMe/r²
r = √GMe/g
r = √(6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)/(0.855 m/s²)
r = 2.16 x 10⁷ m or
r = 21,610 Km
.