Recent content by lavankohsa

OK, I have cleared my confusion by reading university physics book. Thanks all for helping.

And i I have one more confusion. Why we don't include this muscular force of the boy in work -energy theorem.

OK, suppose I lift a book from ground then normal force would do work because point of application moves up. And in above question if ladder does no work while climbing then Boy has to do work by internal force . am I correct ?

Ladder applies force on the boy and boy moves up and the displacement is in direction of force applied. That means ladder does work on the boy.

Ladder doesn't move but does ladder do work on the boy when boy is climbing ? That is I am not able to understand.

It pushes the ladder down.

No, but ladder is involved in going up then it should do work when boy goes up

Homework Statement In a children's park, there is a slide which has a total length of 10 m and a height of 8.0 m . Vertical ladder are provided to reach the top. A boy weighing 200 N climbs up the ladder to the top of the slide and slides down to the ground. The average friction offered by the...

Homework Statement two proton beams going in same direction repel each other whereas two wires carrying currents in same direction attract each other Homework EquationsThe Attempt at a Solution two proton going in same direction then it makes two current going in same direction. And current...

Homework Statement Find the potential difference Va-Vb between points a and b. Homework EquationsThe Attempt at a Solution Each capacitors left plate will have difference amount of positive charge and right plate will have negative charge. After that i don't know how to proceed.

OK thanks . But they should mention it .

Homework Statement Figure shows two rigid vessels A and B, each of volume 200 cm3 containing an ideal gas (Cv = 12.5 J/mol-K). The vessels are connected to a manometer tube containing mercury. The pressure in both the vessels is 75 cm of mercury and the temperature is 300 . (b) 5.0 J of heat...

Ok i solved it i guess. The normal force will be 204 N. Suppose the force by the side wall is F(upward). So for equilibirium, Patm*A+weight of water=N+F. 1*105*20*10-4+5=204+F 205=204+F so F=1N

Will the normal force of base will be 204 N or weight of water which is 5 N ?? I know the arrow should be longer as we move down because pressure increases as we go down.

For first part i found out the pressure at the bottom which is Patm+density*g*h and multiply it by base area. The answer i got 204 N. Now for 2nd part i calculated the weight of water (vol*density) which is 5N. I drew below diagram,