Answer:
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Answer:
153.6 kN
Explanation:
The elastic constant k of the block is
k = E * A/l
k = 95*10^9 * 0.048*0.04/0.25 = 729.6 MN/m
0.12% of the original length is:
0.0012 * 0.25 m = 0.0003 m
Hooke's law:
F = x * k
Where x is the change in length
F = 0.0003 * 729.6*10^6 = 218.88 kN (maximum force admissible by deformation)
The compressive load will generate a stress of
σ = F / A
F = σ * A
F = 80*10^6 * 0.048 * 0.04 = 153.6 kN
The smallest admisible load is 153.6 kN
Answer:
original mass of the block of ice is 38.34 gram
Explanation:
Given data
cup mass = 150 g
ice temperature = 0°C
water mass = 210 g
water temperature = 12°C
ice melt = 2 gram
to find out
solution
we know here
specific heat of aluminum is c = 0.900 joule/gram °C
Specific heat of water C = 4.186 joule/gram °C
so here temperature difference is dt = 12- 0 = 12°C
so here heat lost by water and cup are given by
heat lost = cup mass × c × dt + water mass × C × dt
heat lost = 150 × 0.900 × 12 + 210 × 4.186 × 12
heat lost = 12168.72 J
so
mass of ice melt here = heat lost / latent heat of fusion
here we know latent heat of fusion = 334.88 joule/gram
so
mass of ice melt = 12168.72 / 334.88
mass of ice melt is 36.337554 gram
so mass of ice is here = mass of ice melt + ice melt
mass of ice = 36.337554 + 2
mass of ice = 38.337554 gram
so original mass of the block of ice is 38.34 gram
