A 1.00-kg beaker containing 2.00 kg of oil (density = 916 kg/m3) rests on a scale. A 1.50-kg block of iron is suspended from a spring scale and is completely submerged in the oil (see figure below). Find the equilibrium readings of both scales.
1.00-kg--------------m
2.10 kg ---------------mo
1.50-kg------------mb
pi=7860kg/m^3
po=916 kg/m^3
FB= Vb x po x g
=mb x po x g/pi
=1.50 x916 x 9.8 /7860 = 1.713 N
==================
UPPER
F=mb X g -FB = 1.50 X 9.8 -1.713 N = 12.987 N
==================
LOWER
F=FB +(mb+m)g = 1.713 N +(2.10+ 1.00) x9.8= 32.093 N
A 1.00-kg beaker containing 2.00 kg of oil (density = 916 kg/m3) rests on a scale. A 1.50-kg block of iron is suspended...
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