A 1.0 kg ball is attached to a 1.0 m string and whirled in a horizontal at a con

Question

A 1.0 kg ball is attached to a 1.0 m string and whirled in a horizontal at a constant speed of 3.0 m/s. The work done by the string during each revolution is: (To find the force you can use The kilowatt is a unit for the same quantity as: a) erg/sec; b) electron volt; c) horsepower; d) ft-lb-hour; e) none of these Ten joules of elastic potential energy is stored by compressing a spring. A 2kg object is placed on top of this the spring is projected straight up. If we neglect the energy dissipated in the spring, how high will the object rise? (

Explanation / Answer

Q7.

work done=dot product of force and displacement

as in a circular motion, force is perpendicular to the displacement,

dot product is zero.

hence work done is zero.

Q8. kilowatt is unit of power.

which is same energy per time

hence option A is correct.

Q9 if height of the object is h m,

then potential energy gained by the object=initial potential energy stored in the object

==>2*9.8*h=10

==>h=10/(2*9.8)=0.5102 m

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