Answer
(a) $24.5^{\circ}$
(b) $4.77\times 10^{20}N$
(c) $0.00649m/s^2$
Work Step by Step
(a) We can find the direction of the net force as
$\theta=tan^{-1}(\frac{F_y}{F_x})$
$\theta=tan^{-1}(\frac{1.98\times 10^{29}}{4.34\times 10^{20}})$
$\theta=24.5^{\circ}$
(b) The magnitude of the net force can be determined as follows:
$F=\sqrt{F_x^2+F_y^2}$
We plug in the known values to obtain:
$F=\sqrt{(4.34\times 10^{20})^2+(1.98\times 10^{20})^2}$
$F=4.77\times 10^{20}N$
(c) We can find the required acceleration as
$a=\frac{F}{m}$
We plug in the known values to obtain:
$a=\frac{4.77\times 10^{20}}{7.35\times 10^{22}}$
$a=0.00649m/s^2$
