10 questions · timed · auto-graded
$\therefore$ Total force exerted, f = kx = 150
Here, k is the spring constant of the spring in the bullworker.$\therefore\text{k}=\frac{150}{0.2}=\frac{1500}{2}=750\text{N/m}$
Hence, the spring constant of the spring in the bullworker is 750N/m.
$\text{R}=\sqrt{(48^2+20^2)}$
$=\sqrt{2304+400}$
$=\sqrt{2704}=52\text{N}$
$\therefore$ The magnitude of the total force exerted by the limb on the monkey is 52N.
$\text{F}=\frac{6.67\times10^{-11}\times7.36\times10^{22}\times6\times10^{24}}{3.8\times3.8\times10^{16}}$
$=\frac{6.67\times7.36\times10^{35}}{(3.8)^2\times10^{16}}$
$=20.3\times10^{19}=2.03\times10^{20}$
$\approx2.0\times10^{20}\text{N}$
$\therefore$ The weight of the moon is $2.0\times10^{20}\text{N}.$
$\text{f}_{\text{e}}=\frac{1}{4\pi\in_{0}}\times\frac{\text{q}^2}{\text{r}^2}$
$=\frac{9\times10^9\times(1.6)^2\times10^{-38}}{\text{r}^2} $
Gravitational force between the protons is given by$\text{f}_{\text{g}}=\frac{\text{Gm}^2}{\text{r}^2}$
$=\frac{6.67\times10^{-11}\times(1.67\times10^{-27})^2}{\text{r}^2}$
On dividing fe by fgfe by fg, we get:$\frac{\text{f}_{\text{e}}}{\text{f}_{\text{g}}}=\frac{1}{4\pi\in_0}\times\frac{\text{q}^2}{\text{r}^2}\times\frac{\text{r}^2}{\text{Gm}^2}$
$=\frac{9\times10^9\times1.6\times1.6\times10^{-38}}{6.67\times10^{-11}\times1.67\times1.67\times10^{-54}}$
$=\frac{9\times(1.6)^2\times10^{-29}}{6.67\times(1.67)^2\times10^{-65}}$
$=1.24\times10^{36}$