Force applied on the car due to engine is given as
towards right
Also there is a force on the car towards left due to air drag
towards left
now the net force on the car will be given as
![\vec F_{net} = \vec F_1 + \vec F_2](https://tex.z-dn.net/?f=%5Cvec%20F_%7Bnet%7D%20%3D%20%5Cvec%20F_1%20%2B%20%5Cvec%20F_2)
now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.
So we can say
![F_{net} = F_1 - F_2](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20F_1%20-%20F_2)
![F_{net} = 300 - 150](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20300%20-%20150)
![F_[net} = 150 N](https://tex.z-dn.net/?f=F_%5Bnet%7D%20%3D%20150%20N)
So here net force on the car will be 150 N towards right and hence it will accelerate due to same force.
Answer:
D
Explanation:
I’m pretty sure it’s correct but I don’t really know. Just trying to pass science
Explanation:
We have,
Semimajor axis is ![4\times 10^{12}\ m](https://tex.z-dn.net/?f=4%5Ctimes%2010%5E%7B12%7D%5C%20m)
It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :
![T^2=\dfrac{4\pi ^2}{GM}a^3](https://tex.z-dn.net/?f=T%5E2%3D%5Cdfrac%7B4%5Cpi%20%5E2%7D%7BGM%7Da%5E3)
G is universal gravitational constant
M is solar mass
Plugging all the values,
![T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s](https://tex.z-dn.net/?f=T%5E2%3D%5Cdfrac%7B4%5Cpi%20%5E2%7D%7B6.67%5Ctimes%2010%5E%7B-11%7D%5Ctimes%201.98%5Ctimes%2010%5E%7B30%7D%7D%5Ctimes%20%284%5Ctimes%2010%5E%7B12%7D%29%5E3%5C%5C%5C%5CT%3D%5Csqrt%7B%5Cdfrac%7B4%5Cpi%5E%7B2%7D%7D%7B6.67%5Ctimes10%5E%7B-11%7D%5Ctimes1.98%5Ctimes10%5E%7B30%7D%7D%5Ctimes%284%5Ctimes10%5E%7B12%7D%29%5E%7B3%7D%7D%5C%5C%5C%5CT%3D4.37%5Ctimes%2010%5E9%5C%20s)
Since,
![1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}](https://tex.z-dn.net/?f=1%5C%20s%3D3.17%5Ctimes%2010%5E%7B-8%7D%5C%20%5Ctext%7Byears%7D%5C%5C%5C%5C4.37%5Ctimes%2010%5E9%5C%20s%3D4.37%5Ccdot10%5E%7B9%7D%5Ccdot3.17%5Ccdot10%5E%7B-8%7D%5C%5C%5C%5C4.37%5Ctimes%2010%5E9%5C%20s%3D138.52%5C%20%5Ctext%7Byears%7D)
So, the orbital period of a dwarf planet is 138.52 years.