1. Gold, has a mass of 19.32g for wach cubic centimeter of volume, is most ductile metal and can be pressed into a thin leaf or drawn out into long fiber.

-If a sample of gold, with a mass of 27.63g is pressed into a leaf of 1.000micometers thickness, what is the area of the leaf?
-If instead the gold is drawn out into cylinder fiber of radius 2.500 micrometers, what is the length of the fiber?

thickness is 1.000 micrometers
density=19.32
mass: 27.63

a) area=(27.63)/((19.32)(1.0x10^-4))=1.43g/cm (of leaf)
b) Volume=AxD=1.43x19.32=27.63
27.63=(pie)r^2L
27.63=(pie)(2.500)^2L=1.408 Length of leaf?

2. Time Standards are now based on atomic clocks. A promising second standard is based on pulsars, which are rotating neutron stars (highly compact stars consisting only of neutrons.) Some rotate at a rate that is highly stable, sending out a radio beacon that sweeps briefly across Earth once with each rotation, like a lighthouse beacon. Pulsar PSR 1937+21 is an example; it rotates once every 1.55780644887275 (plus and minus) 3 ms, where trailing (plus and minus) 3 indicates the uncertainty in the last decimal place. (it does not mean plus and minus 3 ms)
a)

To find the area of the leaf, you need to divide the mass of the gold (27.63g) by the density of gold (19.32g/cm³) and the thickness of the leaf (1.000 micrometers or 1.0x10^-4 cm):

Area = (mass) / (density x thickness)
= 27.63g / (19.32g/cm³ x 1.0x10^-4 cm)
= 1.43 cm²

So, the area of the leaf is 1.43 cm².

To find the length of the fiber, you can use the volume of gold, which remains the same as the mass of the gold (27.63 cm³), and the formula for the volume of a cylinder:

Volume = πr²L

Since the radius of the fiber is given as 2.500 micrometers or 2.500x10^-4 cm, you can rearrange the formula to solve for the length (L):

27.63cm³ = π(2.500x10^-4 cm)²L
L = 27.63cm³ / (π(2.500x10^-4 cm)²)
= 1.408 cm

So, the length of the fiber is 1.408 cm.