A 5 m length of strong steel with a cross-sectional area of 20 cm^2 is compresse
ID: 1447799 • Letter: A
Question
A 5 m length of strong steel with a cross-sectional area of 20 cm^2 is compressed with a force of 20,000 N. If the Young’s modulus for this steel is 200 GPa, what is the change in length of the steel?
A gas mixture in a diesel engine is initially held at in a volume of 200 cm^3 at 27 degrees Celsius. The gas is compressed to a volume of 15 cm^3. During the compression, the pressure increases from 1 atmosphere to 50 atmospheres. What is the final temperature?
A small piece of rock is weighed in air and then weighed while immersed in oil of density 800 kg/m^3. If the weight in air is 21.5 N and the weight in the oil is 11.8 N, calculate the density of the rock.
Explanation / Answer
1) young' modulus = stress/strain
Y = F/A/l/L
l = FL/AY
l = 20000*5/(20*10^-4*200*10^9)
l = 0.00025 m
change in length of steel = 0.25 mm
2) Using the equation
P1V1/T1 = P2V2/T2
1*200 / (273+27) = 50*15/T2
T2 = 1125 K
final temperature = (1125-273) = 852oC
3) volume of oil displaced = volume of object
volume of oil displaced = mass* density
weight of oil displaced = reduction in weight when immersed in oil
weight of oil displaced = (21.5-11.8) = 9.7 N
mass = 9.7/9.8 = 0.989 kg
volume = 0.989*800= 791.83 m3
object has weight 21.5 N
so mass = 21.5 /9.8 = 2.2 kg
density = mass/volume = 2.2/791.83 = 0.00277 kg/m3
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