loads of phy questions

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更新1:

3. question: s1353.photobucket/user/s031026/media/P1020770_zpsf08ca013 M in equation v=sqrt(GM/r)/ Kepler's 3rd law is the mass of the object being orbited. When there is no object in the center of rotation
how can the above equations be applied?

更新2:

‘which provides the gravitational force to pull the other star to perform circular motion.’ Do the o stars orbit each other then?

更新3:

5. What do you mean by 'w(= 2.pi/T) is not constant'? that me circular motion under gravity is not an uniform circular motion?

更新4:

6. s1353.photobucket/user/s031026/media/projectilemotion2_zps95e0db57 how to solve this question?

更新5:

7. could you please explain further on this question? I still dun get it... Million thanks!

更新6:

last question 4) normal reaction and weight are always equal to each other
eg. when we are standing on the ground... where does the centripetal force e in this case?

1. Use conservation of momentum
120x0.005 = 1.005v (v is the mon velocity after collision) v = 0.597 m/s hence
height reached = 0.597^2/2g m = 0.0178 m 2. The wer should be option B. 3. Can't see the question. 4. No. It is equal to the difference beeen the weight and normal reaction. 5. In your question
w(= 2.pi/T) is not constant. v^2 = GM/r = r(2.pi/T)^2 This gives r^3 = [GM/4.pi^2].T^2 which is just Kepler's 3rd law 6. Angle (theta) depends on the vertical and horizontal velocity ponents. 7. Net force = (impact force - weight) = rate of change of momentum 8. Difference beeen the weight and normal reaction force. 2013-08-27 20:20:10 补充： Q5: although the question cannot be seen
I think I could still wer your question based on your posted solution. The o stars are orbiting about their "centre of mass". There is NO OBJECT located at the centre of rotation.... 2013-08-27 20:20:43 补充： (cont'd)...The mass of 1x10^30 kg is the mass of a star
which provides the gravitational force to pull the other star to perform circular motion. 2013-08-27 20:21:39 补充： sorry
the above is Q3
not Q5. 2013-08-28 20:02:20 补充： Q3: Kepler's Third Law is not used in this question. The binary star system is clearly shown in the diagram of the question. Each star revolves about the system centre that marked with "x". 2013-08-28 20:08:08 补充： Q5: v = rw is a general equation
but you attempt to apply it to the condition of a stable orbit. For a stable orbit to be establsihed
w decreases (due to lengthening of period T) with increasing r. 2013-08-28 20:23:09 补充： Q6: Horizontal velocity = 10.cos(30) m/s Vertical velocity = 10.sin(30) m/s Hence
use v^2 = u^2 + 2as to find the vertical velocity before hitting the ground v^2 = [10.sin(30)]^2 + 2g(0.3) v = 5.568 m/s 2013-08-28 20:23:27 补充： Use s = ut + (1/2)at^2 to find the time of travel from Y to Z 0 = 5.568t+(1/2)gt^2 t = 1.114 m/s Thus
distance YZ = 10.cos(30) x 1.114 m = 9.643 m 2013-08-28 20:29:11 补充： 7. Weight is always present
thus you need to take it into account The man is acted by 2 force
his weight W and normal reaction R(which is the impact force) Net force = R-W Because -force = rate of change of momentum thus R-W = mv/t R = mv/t + W 2013-08-28 20:31:32 补充： See Q7 in "意见"section. 2013-08-29 23:07:22 补充： See "意见"section for your last question. 2013-08-29 23:15:31 补充： Answer to your "last question". The normal reaction and weight are equal to each other when a person is IN EQUILIBRIUM. When the person is moving in circular motion
he is surely NOT in equilbrium.... 2013-08-29 23:16:18 补充： (cont'd)... As such
the normal reaction and weight are not equal. The difference beeen the o gives the centripetal force.

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