<span>So, if the man weight 900 newtons on Earth then that means, using F=ma, that the mass of the man is approximately 91.84 kg. This is because 900N=m(9.8m/s^2), and so it follows that 900/9.8=91.84. Using the man's found mass we then plug this into F=ma again. It follows that F=(91.84)(25.9)=2378.57N. This means that the man "weighs" 2378.57 Newtons on Jupiter, or about 2.5x as great as his weight on Earth. This makes sense, considering that 25.9/9.8 is approximately equal to 2.64.</span>
If the resistance of the Air is ignored, we can use the theory given by Galileo in which he warned that the thermal velocity of a body in free fall was given by
![v= \frac{1}{2}gt](https://tex.z-dn.net/?f=v%3D%20%5Cfrac%7B1%7D%7B2%7Dgt)
Where
g = Gravitational acceleration
t = time
As we can see the speed of objects in free fall is indifferent to the position that is launched (as long as the resistance of the air is ignored) or its mass.
Both bodies will end with the same thermal speed.
Density <em>ρ</em> is mass <em>m</em> per unit volume <em>v</em>, or
<em>ρ</em> = <em>m</em> / <em>v</em>
Solving for <em>v</em> gives
<em>v</em> = <em>m</em> / <em>ρ</em>
So the given object has a volume of
<em>v</em> = (130 g) / (65 g/cm³) = 2 cm³
Answer:
F = 20 N
Explanation:
It is given that,
The fulcrum is 1 m from the left end, and 3 m from the right end. A block of 60 N is hung to the left end. We need to find the force needed on the right end such that it balances it. The torque acting on the left end is equal to the torque acting on another side. Let F is the force needed on the right end. So,
![60\times 1=F\times 3\\\\F=20\ N](https://tex.z-dn.net/?f=60%5Ctimes%201%3DF%5Ctimes%203%5C%5C%5C%5CF%3D20%5C%20N)
So, the force of 20 N is needed on the right end.
Actual displacement that he required to move
towards North
Displacement that he moved due to snow is
at 47 degree North of East
now in vector component form we can say
![d_1 = 5.4 \hat j](https://tex.z-dn.net/?f=d_1%20%3D%205.4%20%5Chat%20j)
![d_2 = 8.1 cos47 \hat i + 8.1 sin47 \hat j](https://tex.z-dn.net/?f=d_2%20%3D%208.1%20cos47%20%5Chat%20i%20%2B%208.1%20sin47%20%5Chat%20j)
![d_2 = 5.52 \hat i + 5.92 \hat j](https://tex.z-dn.net/?f=d_2%20%3D%205.52%20%5Chat%20i%20%2B%205.92%20%5Chat%20j)
now the displacement that is more required to reach the destination is given as
![d = d_1 - d_2](https://tex.z-dn.net/?f=%20d%20%3D%20d_1%20-%20d_2)
![d = 5.4\hat j - (5.52 \hat i + 5.92\hat j)](https://tex.z-dn.net/?f=d%20%3D%205.4%5Chat%20j%20-%20%285.52%20%5Chat%20i%20%2B%205.92%5Chat%20j%29)
![d = -5.52 \hat i - 0.52 \hat j](https://tex.z-dn.net/?f=d%20%3D%20-5.52%20%5Chat%20i%20-%200.52%20%5Chat%20j)
so the magnitude of the displacement is given as
![d = \sqrt{5.52^2 + 0.52^2}](https://tex.z-dn.net/?f=d%20%3D%20%5Csqrt%7B5.52%5E2%20%2B%200.52%5E2%7D)
![d = 5.54 km](https://tex.z-dn.net/?f=d%20%3D%205.54%20km)
its direction is given as
![\theta = tan^{-1}\frac{0.52}{5.52} = 5.38 degree](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%5Cfrac%7B0.52%7D%7B5.52%7D%20%3D%205.38%20degree)
so it is 5.54 km towards 5.38 degree North of West.