Asked by Tina
A beaker of water placed on a scale produces a reading of 0.910kg. An object of mass 0.27kg and volume 9.40 x 10^-5m^3 is lowered into the water by a string. By how much will the reading on scale change?
Answers
Answered by
Anonymous
first check the density of this object.
.27 kg/.000094 m^3 = 2872 kg/m^3
water density is 1000 so this thing sinks.
The tension in the string will decrease by the volume of the object * density of water * g
= .000094 * 1000 g = .094 g Newtons
the water pushes up with .094 g so the scale has to push up with an additional .094 g N
since our scale is calibrated for mass on earth in kg it will read an additional .094 kg
.27 kg/.000094 m^3 = 2872 kg/m^3
water density is 1000 so this thing sinks.
The tension in the string will decrease by the volume of the object * density of water * g
= .000094 * 1000 g = .094 g Newtons
the water pushes up with .094 g so the scale has to push up with an additional .094 g N
since our scale is calibrated for mass on earth in kg it will read an additional .094 kg
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