A 1m^3 cubic block of wood of specific gravity 0.7 floats on water. Note: specif
ID: 1410010 • Letter: A
Question
A 1m^3 cubic block of wood of specific gravity 0.7 floats on water. Note: specific gravity = rho_substanced rho_water How much of it is immersed? How many 60 kg swimmers can stand on top of the block before it gets completely immersed in water? An object "weighs" 10 lb in the air and 6 lb in water, a. What is the specific gravity b. How much would it "weigh" in oil of specific gravity 0.6 [sg = 2.5] A cube of wood 10 cm on a side and density 0.8 g/cm^3 sits in oil and water. How much of the block is in oil and how much is in water if the oil has a density of 0.65 g/cm^3? [571 cc's in oil]Explanation / Answer
2 ans
Volume of block V=1m^3
Specific gravity =>s.g=0.7
This concept belongs mechanical properties of fluids
According to the principle of floatation
(A)
Volume of body in side the liquid =[s.g]V
=0.7×1=0.7m^3
(B)
W=n V gh(1-s.g)
60=n×1×9.8×1(1-0.7)
n=20.41
The swimmer's is stand on the block =20.4
(3) ans
The weight of body in air =>W1=10 in
The weight of body in water=>W2=6ib
This concept belongs mechanical properties of fluids
According to Archimedes principle
(A)
The specific gravity =w1/(w1-w2)
=10/(10-6)=2.5
(B)
The specific gravity =>s.g=0.6
Now we find weight of the body in oil w3
So specific gravity=(w1-w3)/(w1-w2)
=>0.6=10-w3/(10-6)
=>2.4=10-w3
w3=7.6ib
(4) ans
Density of the water dw=1g/cm^3
Density of the oil. do= 0.65 g/cm^3
This concept belongs mechanical properties of fluids
According to principle of flotation
The volume body in two different liquids
Vo/Vw = dw/do =1/0.65
Vo/Vw =1/0.65
The volume body in oil=1 m3
The volume body in water =0.65 m^3
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.