Answer:
Acceleration a=10.234 m/s²
Direction α=85.3°
Explanation:
Given data
F₁=1.20×10⁻⁵N
F₂=0.500×10⁻⁶N
mass m=6.00×10⁻⁷kg
According to Newton second law.
Fnet=∑F
[tex]F_{net}=F_{1}+F_{2}+W=ma\\[/tex]
Substitute the given values
[tex]a=\frac{F_{1}+F_{2}+W}{m} \\a=\frac{1.20*10^{-5}j+(0.500*10^{-6} )i+(mg)}{m}\\a=\frac{1.20*10^{-5}j+(0.500*10^{-6} )i+(6.00*10^{-7}(-9.8m/s^{2} ))}{6.00*10^{-7}}\\a=[10.2j+0.833i]m/s^{2}[/tex]
So the magnitude of acceleration is given by
[tex]a=\sqrt{a_{x}^{2}+a_{y}^{2} } \\a=\sqrt{(10.2)^{2}+(0.833)^{2} }\\a=10.234m/s^{2}[/tex]
For the direction is given by:
[tex]\alpha =tan^{-1}(\frac{a_{y} }{a_{x}} ) \\\alpha=tan^{-1}(\frac{10.2 }{0.833} )\\\alpha =85.3^{o}[/tex]
