The diameter of the circle is 18 m. Eugene incorrectly says that the circumference of the circle is about 113.04 m. What mistake did Eugene make? Use 3.14 for pi.
Answer:
(a) 1767.43 N
(b) 182.45 N
Explanation:
Radius of earth, R = 6450 km
Weight of person, W = 7070 N
mass of person, m = W / g = 7070 / 9.8 = 721.4 kg
(a) h = 6450 km
The value of acceleration due to gravity on height is given by


g' = g / 4 = 9.8 / 4 = 2.45 m/s^2
The weight of the person at such height is
W' = m x g' = 721.4 x 2.45
W' = 1767.43 N
(b) h = 33700 km
The value of acceleration due to gravity on height is given by


g' = g x 0.0258 = 9.8 x 0.0258 = 0.253 m/s^2
The weight of the person at such height is
W' = m x g'
W' = 721.4 x 0.253
W' = 182.45 N
Answer:
Explanation:
Apply the law of conservation of energy

![Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)](https://tex.z-dn.net/?f=Gm_1m_2%5B%5Cfrac%7B1%7D%7Br_f%7D%20-%5Cfrac%7B1%7D%7Br_1%7D%20%5D%3D%5Cfrac%7B1%7D%7B2%7D%20%28m_1v_1%5E2%2Bm_2v_2%5E2%29)
from the law of conservation of the linear momentum

Therefore,
![Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)](https://tex.z-dn.net/?f=Gm_1m_2%5B%5Cfrac%7B1%7D%7Br_f%7D%20-%5Cfrac%7B1%7D%7Br_1%7D%20%5D%3D%5Cfrac%7B1%7D%7B2%7D%20%28m_1v_1%5E2%2Bm_2v_2%5E2%29)
![=\frac{1}{2} [m_1v_1^2+m_2[\frac{m_1v_1}{m_2} ]^2]\\\\=\frac{1}{2} [m_1v_1^2+\frac{m_1^2v_1^2}{m_2} ]\\\\=\frac{m_1v_1^2}{2} [\frac{m_1+m_2}{m_2} ]](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%7D%20%5Bm_1v_1%5E2%2Bm_2%5B%5Cfrac%7Bm_1v_1%7D%7Bm_2%7D%20%5D%5E2%5D%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D%20%5Bm_1v_1%5E2%2B%5Cfrac%7Bm_1%5E2v_1%5E2%7D%7Bm_2%7D%20%5D%5C%5C%5C%5C%3D%5Cfrac%7Bm_1v_1%5E2%7D%7B2%7D%20%5B%5Cfrac%7Bm_1%2Bm_2%7D%7Bm_2%7D%20%5D)
![v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]](https://tex.z-dn.net/?f=v_1%5E2%3D%5B%5Cfrac%7B2Gm_2%5E2%7D%7Bm_1%2Bm_2%7D%20%5D%5B%5Cfrac%7B1%7D%7Br_f%7D%20-%5Cfrac%7B1%7D%7Br_1%7D%20%5D)
Substitute the values in the above result
![v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]](https://tex.z-dn.net/?f=v_1%5E2%3D%5B%5Cfrac%7B2Gm_2%5E2%7D%7Bm_1%2Bm_2%7D%20%5D%5B%5Cfrac%7B1%7D%7Br_f%7D%20-%5Cfrac%7B1%7D%7Br_1%7D%20%5D)
![=[\frac{2(6.67\times 10^-^1^1)(107)^2}{27+107} ][\frac{1}{26} -\frac{1}{41}] \\\\=1.6038\times 10^-^1^0\\\\v_1=\sqrt{1.6038\times 106-^1^0} \\\\=1.2664 \times 10^-^5m/s](https://tex.z-dn.net/?f=%3D%5B%5Cfrac%7B2%286.67%5Ctimes%2010%5E-%5E1%5E1%29%28107%29%5E2%7D%7B27%2B107%7D%20%5D%5B%5Cfrac%7B1%7D%7B26%7D%20-%5Cfrac%7B1%7D%7B41%7D%5D%20%5C%5C%5C%5C%3D1.6038%5Ctimes%2010%5E-%5E1%5E0%5C%5C%5C%5Cv_1%3D%5Csqrt%7B1.6038%5Ctimes%20106-%5E1%5E0%7D%20%5C%5C%5C%5C%3D1.2664%20%5Ctimes%2010%5E-%5E5m%2Fs)
B) the speed of the sphere with mass 107.0 kg is

\\\\=3.195\times 10^-^6m/s](https://tex.z-dn.net/?f=%3D%5B%5Cfrac%7B27%7D%7B107%7D%20%5D%281.2664%20%5Ctimes%2010%5E-%5E5%29%5C%5C%5C%5C%3D3.195%5Ctimes%2010%5E-%5E6m%2Fs)
C) the magnitude of the relative velocity with which one sphere is

D) the distance of the centre is proportional to the acceleration

Thus,

and

When the sphere make contact with eachother
Therefore,

And

The point of contact of the sphere is

No force is required to lift that balloon. In fact, force is required to hold it down, and if you let go, it's up, up, and away.
Since the balloon's density is less than the density of the air around it, it's lighter than the air it displaces, there is a net upward buoyant force acting on it, and it floats up !
Answer:
physic is branch of science in which math is its brother and will deal with the equation,law and evidence of natural phenomenon.