Question

You purchase a rectangular piece of metal that has dimensions 10.0 mm × 10.0 mm × 25.0 mm and mass 1.92×10−2 kg . The seller tells you that the metal is gold. To check this, you compute the average density of the piece. What value do you get? Were you cheated? yes or no 6 of 15 Constants You are designing a hydraulic lift for an automobile garage. It will consist of two oil-filled cylindrical pipes of different diameters. A worker pushes down on a piston at one end, raising the car on a platform at the other end. (See (Figure 1).) To handle a full range of jobs, you must be able to lift cars up to 3000 kg , plus the 550 kg platform on which they are parked. To avoid injury to your workers, the maximum amount of force a worker should need to exert is 100 N . What should be the diameter of the pipe under the platform? If the worker pushes down with a stroke 60 cm long, by how much will he raise the car at the other end?

Answer 1

(a)

the volume of the metal is the product of the 3 dimensions; converting all measurements to meters gives us

Volume = 10x10^-3 m x 10 x 10^-3 m x 10 x 10^-3 m

=1 x10^-6 m^3

if its mass is 1.92*10^-2kg, then

density = mass/volume = 1.92*10^-2kg/1x10^-6 m^3 = 19200kg/m^3

this is remotely close to the density of gold which is 19320 kg/m^3. So I was not cheated.

Hope it was helpful..

According to Chegg policy I'm supposed to answer only first question when multiple questions care asked.. so I have answered only first question.