An object is solid throughout. When the object is completely submerged in ethyl alcohol, its apparent weight is 15.1 N. When completely submerged in water, its apparent weight is 12.7 N. What is the volume of the object?
2007-07-10 10:14:47 · 3 answers · asked by Eunice C 1 in Science & Mathematics ➔ Physics
It helps to have the density of EtOH as info. For simplicity sake, call it 0.8 g/cc.
In EtOH apparent wt = actual wt (AW) - V *density displaced.
15.1 = AW - 0.8 V
In water, the same holds true
12.7 = AW - V.
Then 2.4 = 0.2 V and V= 12 cc
Sneaky, hah?
2007-07-10 10:25:24 · answer #1 · answered by cattbarf 7 · 0⤊ 1⤋
obvious weight = weight in air - buoyant stress. 4.7=25-facebook, facebook=20.3 N. facebook= density of water x gravity x quantity. 20.3=one thousand x 9.8 x Vol. Vol=2.071 x 10^-3. Fg=25. Mg=25. Mx9.8=25. M=2.55Kg. Density =Mass/quantity. 2.fifty 5/2.071 x 10^-3=1231.7kg/m^3.
2016-12-14 05:02:53 · answer #2 · answered by lacuesta 4 · 0⤊ 0⤋
12.7 ÷ 9.81 = 1.3kg
1kg of water = 1 L
1.3 kg = 1.3 Litres = 1,300 cc = Volume of object.
2007-07-10 12:33:36 · answer #3 · answered by Norrie 7 · 0⤊ 0⤋