A piece of lead with a mass of 103.5g has a length of 50.00 mm and a height of 34.00 mm. What is the depth

Answer 1

lead density = 11350 kg/m^3

V = 0.1035 / 11340 = 9.126 .10^(-6) m^3

we know that

- L = 50 mm = 0,05 m
- H = 34 mm = 0.034 m

D ?

V = LHD => D = V/(LH)

D = [ 9.126 .10^(-6) ]/[0.05 x 0.034 ] = 5.368 .10^(-3) = 5.368 mm

Answer 2

You must use the density of lead to solve this one. D = 11.34 g/cm^3

D = m/v, so v = m/D

(103.5 g) / (11.34 g/cm^3) = v = 9.13 cm^3

Now do a little geometry and remember that v = l x w x h.
Area = l x w
therefore v = A x h (or thickness)

The area of the lead is (convert mm to cm by dividing by 10 since density is measured in g/cm^3)

5.000 cm x 3.400 cm = 17.00 cm^2

Since v = A x h substitute the calculated volume and area into the equation:

9.13 cm^3 = (17.00 cm^2) (height)
rearrange the equation:

9.13 cm^3 / 17.00 cm^2 = height = 0.5371 cm thick

Answer 3

ok find out its densty (look it up somewhere)

Then use this fomulra.

D = M/V

D = 0.1035 Kg / (0.05 M x 0.034 M x depth (in metres))

depth (in metres) = 0.1035 Kg /( 0.05 M x 0.034 M x denisty of lead).