Question

9. An object of mass 0.5 kg is whirled at the
end of a string 0.8 m long. If the string makes
three revolutions in 1.2 s, find the tension in
the string
(Ans. 35.6 N)​

Answer 1

Answer:

tension = sum of centrietal force+gravity

notice these are vectors.

tensionvertical= mg down

tensionhorizonal= m w^2 * rad

= .5*(2PI/.4)^2 * .8

now, after calculating that, you add them as vectors

tension= sq rt ( (mg)^2+(horizontal above)^2 )

= 35.6 N

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