The correct formula to use for the situation given above is: F = MA, where F is the applied force, M is the mass of the object and A is the acceleration.
From the details given in the question, we are told that:
F = 18, 400N
M = 145 g = 145 / 1000 = 0.145 kg
A = ?
From the equation F = MA
A = F / M
A = 18,400 / 0.145 = 126,896.55 = 1.27 *10^5.
Therefore, the correct option is C.
The sharp nail has a less surface area in comparison to a blunt nail and pressure is inversely proportional to area so it is easier to Hamer a sharp nail into a wood rather than having a blunt nail in wood
<span>Discrimination is illegal, but caste system is legal.
So answer: False</span>
1. 0.16 N
The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:
![F=G\frac{Mm}{r^2}](https://tex.z-dn.net/?f=F%3DG%5Cfrac%7BMm%7D%7Br%5E2%7D)
where
G is the gravitational constant
is the mass of the asteroid
m = 100 kg is the mass of the man
r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid
Substituting, we find
![F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N](https://tex.z-dn.net/?f=F%3D%286.67%5Ccdot%2010%5E%7B-11%7Dm%5E3%20kg%5E%7B-1%7D%20s%5E%7B-2%7D%29%5Cfrac%7B%288.7%5Ccdot%2010%5E%7B13%7D%20kg%29%28110%20kg%29%7D%7B%282000%20m%29%5E2%7D%3D0.16%20N)
2. 1.7 m/s
In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force
![m\frac{v^2}{r}=G\frac{Mm}{r^2}](https://tex.z-dn.net/?f=m%5Cfrac%7Bv%5E2%7D%7Br%7D%3DG%5Cfrac%7BMm%7D%7Br%5E2%7D)
where v is the minimum speed required to stay in orbit.
Re-arranging the equation and solving for v, we find:
![v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cfrac%7BGM%7D%7Br%7D%7D%3D%5Csqrt%7B%5Cfrac%7B%286.67%5Ccdot%2010%5E%7B-11%7D%20m%5E3%20kg%5E%7B-1%7D%20s%5E%7B-2%7D%29%288.7%5Ccdot%2010%5E%7B13%7D%20kg%29%7D%7B2000%20m%7D%7D%3D1.7%20m%2Fs)
<span>A gymnast with mass m1 = 43 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 = 115 kg and length L = 5 m. Each support is 1/3 of the way from each end. Initially the gymnast stands at the left end of the beam.
1)What is the force the left support exerts on the beam?
2)What is the force the right support exerts on the beam?
3)How much extra mass could the gymnast hold before the beam begins to tip?
Now the gymnast (not holding any additional mass) walks directly above the right support.
4)What is the force the left support exerts on the beam?
5)What is the force the right support exerts on the beam?</span>