If the cord has negligible mass and acceleration is finite then the net force on the cord must be zero (ma = 0). Therefore tension on the two ends will be equal and opposite.
yes, m.o.i. Of hemisphere will be same along axis through the centre and the one perpendicular to it. If you want to find the m.o.i about the axis along the surface through its centre then you have to either see the symmetry or use the long integration method.
I have via done integration also...
1. I can't think of anybody having same moment inertia about any axis.
2. a). Sphere
b).Hemisphere also has same m.o.i about any axis passing through the centre of its plane surface.
C). Square sheet.
About any axis in its plane and through its centre.
if you have done SHM then:
T=2pi(m/k)^1/2
From the graph T=0.69sec.
After putting the i am getting k=1.39 N/m.
Am I correct.
Can u tell how to upload picture(from mobile). I am new here
i am just going to give my exams after two months. After that i will be free for 2-3 months, before i join any other institute.
I have done classical physics in detail in past two years.(not of engineering level, as i still hav to join a college)
i like physics very much.
Quantum world looks...
it depends on whether the ball is conducing or non-conducting.
If the ball is conducting then your approach is right. And answer will be in c/m^2.
But as you have mentioned that answer of charge density has been asked in c/m^3 ball is assumed to be non-conducting. Charge densities at inner...
if I'm not understanding wrong, we have to find the 'relative' time period of the planet w.r.t earth. Here relative means : let car A & B start from pt. P in diff. Circles with diff. Speeds. Now relative time period will be that when both will reach pt. Again simultans. To do this problem u can...
i am not saying that the reaction will be sn2, i was just replying to the-exiled's explanation that due to sn2 rxn confg. Can change (of the attacking species)
about the optical activity, i want to confirm if the two solutions (with diff. Compounds) have same confg. (R or S) and same concentrations then will their ability (degree of rotation) to rotate the plane polarised will be same or not.
according to mechanism, carbanion will be the leaving group, as the carbanion will be tetrahedral, therefore i think the confg. Must not change.
But the exam in which i faced this ques, the answer was the other way.
According to them confg. Can change, i want to know how confg. Can change...
There is a compound 1-fluoro,1-chloro,1-bromo, propanone (a substituted ketone).
If we react it with an alkali, we know alkali will attack carbonyl carbon and chloro-fluoro-bromo- methyl carbanion will be the leaving group,
finally acetic acid and chloro-fluoro-bromo- methane will be formed...