A block of copper with a density of 8920 kg/m3, a length of 0.6 m, a width of 0.
ID: 1773985 • Letter: A
Question
A block of copper with a density of 8920 kg/m3, a length of 0.6 m, a width of 0.25 m, and a height of 0.4 m has a string of negligible weight tied to it and is suspended from a force probe.
(a) What is the tension in the string when the block is in the air? N
(b) What is the tension in the string when the block has been lowered into a tank of glycerin (which has a density of 1260 kg/m3) and is completely submerged?
A toy balloon, which has a mass of 3.90 g before it is inflated, is filled with helium (with a density of 0.180 kg/m3) to a volume of 8500 cm3. What is the minimum mass that should be hung from the balloon to prevent it from rising up into the air? Assume the air has a density of 1.29 kg/m3. g
Explanation / Answer
In equiliobrium Fnet = 0
T - m*g = 0
T = m*g
mass m = density *volume = density*length*width*height
Tension T = 8920*0.6*0.25*0.4*9.8 = 5225 N
(b)
buoyancy force Fb = density of glycerine * volume immersed*g
In equilibrium
Fnet = 0
T + Fb - mg = 0
T = mg - Fb
T = (8920*0.6*0.25*0.4*9.8) - (1260*0.6*0.25*0.4*9.8)
Tension T = 4504.08 N
===============================
volume of balloon V = 8500 cm^3 = 8500*10^-6 m^3
buoyancy force Fb = density of air*V*g
mass of helium in balloon m1 = density of helium*volume of balloon = 0.18*8500*10^-6 = 0.00153 kg
mass of balloon m2 = 3.9 g = 3.9*10^-3 = 0.0039 kg
Mass to be hung = M
In equilibrium
Fnet = 0
Fb - (m1+m2)g - Mg = 0
M = (Fb - (m1+m2)g)/g
M = ((1.29*8500*10^-6*9.8)-( 0.00153+0.0039)*9.8)/9.8
M = 0.005535 kg <<<<-------ANSWER
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