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
Answer:
The volume of the balloon increases in the upper atmosphere.
Explanation:
p1= 1 atm
p2= 0.15 atm
V1= 15.6 L
V2= ?
p1*V1= p2 * V2
V2= (p1/p2)*V1
V2= 104 L
Answer:
2.5m/s^2
Explanation:
Step one:
given
distance = 20meters
time = 2 seconds
initial velocity u= 0m/s
let us solve for the final velocity
velocity = distance/time
velocity= 20/2
velocity= 10m/s

divide both sides by 40

Can neither be created or destroyed.