C) it decreased
because atmospheric pressure decreases as we move up.
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,
![=\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
![=[\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
Answer:
(a) 1.21 m/s
(b) 2303.33 J, 152.27 J
Explanation:
m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s
(a) Let their velocity after striking is v.
By use of conservation of momentum
Momentum before collision = momentum after collision
m1 x u1 + m2 x u2 = (m1 + m2) x v
- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v
v = ( - 356.25 + 607.94) / 208 = 1.21 m /s
(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2
= 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)
= 0.5 (1335.94 + 3270.7) = 2303.33 J
Kinetic energy after collision = 1/2 (m1 + m2) v^2
= 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J
We shall consider two properties:
1. Temperature difference
2. Thermal conductivity of the material
Use a cylindrical rod of a given material (say steel) which is insulated around its circumference.
One end of the rod is dipped in a large reservoir of water at 100 deg.C and the other end is dipped in water (with known volume) at 40 deg. C. The cold water if stored in a cylinder which is insulated on all sides. A thermometer reads the temperature of the cold water as a function of time.
This experiment will show that
(a) heat flows from a region of high temperature to a region of lower temperature.
(b) The thermal energy of a body increases when heat is added to it, and its temperature will rise.
(c) The thermal conductivity of water determines how quickly its temperature will rise. If mercury replaces water in the cold cylinder, its temperature will rise at a different rate because its thermal conductivity is different.
no it can't do this why because I think that it is water and it can not go any where.
