Answer
(a) $\rho=3.7321\times10^{18}kg/m^3$
(b) $W=1.64439\times10^{12}lb$
Work Step by Step
Mass of a neutron star $M=2.7\times10^{28}kg$
$r=1.2\times10^{3}m$
Thus:
$V=\frac{4}{3}\pi r^3$
$V=\frac{4}{3}\pi (1.2\times10^{3}m)^3$
$V=7.23456\times10^{9}m^3$
(a) Density
Density of the neutron star will be $\rho=\frac{M}{V}$
$\rho=\frac{2.7\times10^{28}kg}{7.23456\times10^{9}m^3}$
$\rho=0.37321\times10^{19}kg/m^3$
$\rho=3.7321\times10^{18}kg/m^3$
(b)
Volume of dime $V=2.0\times10^{-7}m^3$
$\rho=3.7321\times10^{18}kg/m^3$
$\rho=\frac{M}{V}$
$M=\rho V$
$M=3.7321\times10^{18}kg/m^3\times2.0\times10^{-7}m^3$
$M=7.4642\times10^{11}kg$
$W=Mg$
$W=7.4642\times10^{11}kg\times9.8m/s^2$
$W=73.14916\times10^{11}N$
Since $1N=0.2248lb$
$W=73.14916\times10^{11}\times0.2248lb$
$W=16.4439\times10^{11}lb$
$W=1.64439\times10^{12}lb$
