Answer: 16J
Explanation:
K.E = m
K.E = X 2 X 16
K.E = 16
Eight joules of work is required to stretch a spring 0.5 meter from its natural length. Find the work required to stretch the sp
1 answer:
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mass of iron block given as
density of iron block is
now the volume of the iron piece is given as
Now when this iron block is complete submerged in oil inside the beaker the buoyancy force on the iron block will be given as
here we know that
= density of liquid = 916 kg/m^3
Now for the reading of spring balance we can say the spring force and buoyancy force on the block will counter balance the weight of the block at equilibrium
So reading of spring balance will be 16.45 N
Now for other scale which will read the normal force of the surface we can write that normal force on the container will balance weight of liquid + container and buoyancy force on block
So the other scale will read 36.47 N