Answer:
penetration probe
Explanation:
it is a penetration probe
To solve this problem, separate it into chunks that you know. You know that there are 2.54 centimeters in 1 inch. You know that there are 100 centimeters in 1 meter. You know that there are 1000 meters in a kilometer. Therefore, we'll convert in this order: 1) from kilometers to meters, 2) from meters to centimeters, and 3) from centimeters to inches.
1) 1km × 1000m/1km
= 1000m
2) 1000m × 100cm/1m
= 100000cm
3) 100000cm × 1in/2.54cm
≈ 39370in
So, there are approximately 39370 inches in a kilometer.
The molecular weight of Mg(OH)2 : 58 g/mol
<h3>Further explanation</h3>
Given
Mg(OH)2 compound
Required
The molecular weight
Solution
Relative atomic mass (Ar) of element : the average atomic mass of its isotopes
Relative molecular weight (M) : The sum of the relative atomic mass of Ar
M AxBy = (x.Ar A + y. Ar B)
So for Mg(OH)2 :
= Ar Mg + 2 x Ar O + 2 x Ar H
= 24 g/mol + 2 x 16 g/mol + 2 x 1 g/mol
= 24 + 32 + 2
= 58 g/mol
Answer:
Energy sources do not have 100% efficiency because <em>the processes of energy conversion to usable forms involves energy losses. </em>
Some have lower efficiencies due to; <u>energy losses in form of heat</u> during conversion, <u>poor technology applied during conversion</u> of energy and<u> lack of desire equipment</u> to use in the energy conversion system.
Explanation:
The desired form of energy for use is derived from conversion of energy from the source using an energy converter into another form which is usable. The efficiency of the energy converter is calculated as;
л = output energy/input energy
The efficiency of energy is limited to the cost of equipment required for conversion from energy source by the energy converter to a form which is usable. Additionally, because energy sources are scarce, the technology to use in energy conversion is a factor affecting energy efficiency in that high efficiency will require advanced technology with better equipment leading higher costs of that energy form. when heat losses are involved during energy conversion, efficiency lowers, thus its better if such losses are used as energy input in another system.
