Recent content by ash4741

Thanks everyone

So if the mass of water displaced= mass of ice then: density of water * volume of water displace = density of ice * volume of ice So (Volume of water displaced)/(volume of ice) =1 and you would subtract 1 from density of ice/ density of water Is that right?

So the answer is 0.931 for the ratio that is exposed? I am just guessing because what I thought was the portion under water is 1.119.

I though that the answer to this division (which is actually 1.119) was the volume of the iceberg that is under water. Sorry, I do not quite understand.

So the 1.101 is the mass of the iceberg? and the mass of the water displaced? But I thought that was the volume of the iceberg that is under the water.

Homework Statement The density of ice is 920kg/m^3, and the density of sea water is 1030kg/m^3. What is the fraction of the total volume of an iceberg that is exposed? given: density sea water = 1030kg/m^3 density ice = 920kg/m^3 Homework Equations density water/density ice...

Thanks ^^

W=1/2mVf^2-1/2mVi^2 W=1/2(69)(0)-1/2(69)(121) W=-4174.5 J

W=1/2m(Vf^2+Vi^2) W=1/2(69)(0^2+11^2) W=1/2(69)(121) W=34.5(121) W=4174.5 This is wrong answer.

Wall excerts work on skater: Please help :( Homework Statement A 69.0-kg short-track ice skater is racing at a speed of 11.0 m/s when he falls and slides into a padded wall that brings him to rest. Assuming that he does not lose any speed during the fall or while sliding across the ice, how...