Answer:
1. Two materials that can be scratched by an iron nail are <u>zinc</u> and <u>aluminum</u>
2. A substance that can be scratched by an iron nail but not a penny is;
Nickel
Explanation:
The hardness of a material can be defined as its ability to resist being plastically deformed at a location due to mechanical scratching, abrasion or indentation
The Mohs hardness scale provides a ranking of materials in the order of hardness with a material having a higher Mohs hardness number being able to scratch other materials which have a lower Mohs hardness number
1. From the Mohs scale of hardness, iron which has an harness number of 4.5 can scratch zinc which has an harness number of 2.5, and iron can also scratch aluminum which has an hardness number of 2.5 to 3
2. A penny is made from copper plated material and will have an outer layer hardness of copper which has an hardness number of 3 on the Mohs scale
Therefore, a penny cannot scratch a material mode of nickel which has an hardness number of 4, while iron which has an hardness of 4.5 will scratch a nickel material
Answer:
5 years
Explanation:
The insurances mandate states that every driver or owner of a car has to have an insurance at least covering third parties damages and hospital care, this is to prevent people from going broke because of a car accident, if you faile to have one they would make you have proof of financial responsability for a year, if ater that failing, you are caught again without insurance withint 5 years the proof of financial responsability increases to 5 years.
Answer:
Explanation:
THE GIVEN sheet can be taken as two horizontal force with surface charge density is
at one surface is ∈_1 =
at oher surface is ∈_2=
the magnitude of electric field due to surface charge is given as
So, electric field at P (2 CM below from surface is) = E_1 +E_2
Answer:
100°C
Explanation:
The heat gained by the ice equals the heat lost by the steam, so the total heat transfer equals 0.
Heat lost by the steam as it cools to 100°C:
q = mCΔT
q = (3 kg) (2.00 kJ/kg/K) (100°C − 120°C)
q = -120 kJ
Total heat so far is negative.
Heat lost by the steam as it condenses:
q = -mL
q = -(3 kg) (2256 kJ/kg)
q = -6768 kJ
Heat absorbed by the ice as it warms to 0°C:
q = mCΔT
q = (6 kg) (2.11 kJ/kg/K) (0°C − (-40°C))
q = 506.4 kJ
Heat absorbed by the ice as it melts:
q = mL
q = (6 kg) (335 kJ/kg)
q = 2010 kJ
Heat absorbed by the water as it warms to 100°C:
q = mCΔT
q = (6 kg) (4.18 kJ/kg/K) (100°C − 0°C)
q = 2508 kJ
The total heat absorbed by the ice by heating it to 100°C is 5024.4 kJ.
If the steam is fully condensed, it loses a total of -6888 kJ.
Therefore, the steam does not fully condense. The equilibrium temperature is therefore 100°C