2- The density of ice is 917 kg/m 3 , and the density of sea water is 1025 kg/m
ID: 2245402 • Letter: 2
Question
2- The density of ice is 917 kg/m3, and the density of sea water is 1025 kg/m3. A swimming polar bear climbs onto a piece of floating ice that has a volume of 4.59 m3. What is the weight of the heaviest bear that the ice can support without sinking completely beneath the water?
Explanation / Answer
Given
Area of a lid, A = 1.20 *10^-2m^2
pressure, p = 9.50 *10^4 Pa
Magnitude of Force,F = P *A
= (9.50 *10^4 Pa) *(1.20 *10^-2m^2)
F = 1140 N
2)
2)Density of ice = 917 kg/m^3
Volume of ice = 4.59 m^3
Mass of ice = Volume x Density
= 917 x 4.59 = 4209.03 kg,,,,,,,,,,,,,,
Weight of ice = mass x g =4209.03 x 9.8 = 41248.494 N................
Density of sea water = 1025 kg/m^3.......
When the bear is on top, the WHOLE ice will b submerged,
meaning the whole volume of the ice is underwater.
Buoyant force = V(d)(g) ,
where V is the volume of ice,
d is the density of the SEA WATER,
g is the gravitational acceleration.
Buoyant force = (4.59 )(1025)(9.8) =46106.55 N,,,,
The weight of the bear = Buoyant force - Weight of ice =
46106.55N - 41248.494N =4858.056 N,,,,,,,,,,,,,,,,,,
Mass of bear = Weight of bear / 9.8 = 4858.056 / 9.8 = 495.72kg
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