A small plastic ball (similar to a ping-pong ball) with mass 2.5g is attached by a thread to the bottom of a beaker. When the beaker is filled with water so that the ball is totally submerged, the tension in the thread is 0.03N. The density of water is 1000kg/m^3 and the acceleration of gravity is 9.81m/s^2.
Determine the diameter of the ball.
2007-07-16 19:49:49 · 3 answers · asked by Doodle 2 in Science & Mathematics ➔ Physics
Mass of the ball 2.5g gives a vertical downward force of
2.5g * 9.81 m/sec^2 = 0.024525N
The upward tension is 0.03N
Add the two forces to get the upward force due to water displacement: 0.03N + 0.0245N = 0.0545N
Mass of displaced water = 0.0545N / 9.81 m/s^2 = 5.558g
Volume of ball = volume of displaced water = 5.558cm^3
Radius of ball = (3V / 4pi)^0.333 = 1.1cm
Diameter = 2 x radius = 2.2cm
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2007-07-16 20:54:49 · answer #1 · answered by Ernst S 5 · 0⤊ 0⤋
T+ mg = [4/3] Ï r^3 d g. = weight of equal volume of water
r^3 =3 [T + mg ] / [4 Ï d g ]
Given T = 0.03 N = [0.03 /g] g = 0.00306 g
r^3 =3 [0.003058 + 0.0025] / [4 Ï 1000 ]
r =0. 01099 m
Diameter = 2.197 cm
2007-07-17 07:23:54 · answer #2 · answered by Pearlsawme 7 · 0⤊ 0⤋
m = F/a = 0.3/9.81 = 0.0306 kg
(this is the mass of the water displaced)
0.0306 kg / 1000 kg-m^-3 = 3.06E-5 m^3 = 30.6 cm^3
2007-07-17 02:59:55 · answer #3 · answered by gebobs 6 · 1⤊ 1⤋