HELPPPPPPPPPPPPPPPPPPPQuestion 17 of 25
2 answers:
You might be interested in
To solve this problem we will start by considering how to calculate the apparent weight. On the sphere this will then be given that the real weight is the sum of the apparent weight and the Buoyant Force. Therefore we will have to
Here
= True Weight
= Apparent Weight
= Buoyant Force
If we seek to find the apparent weight we will have to,
Remember that
V = Volume (Volume Sphere)
= Density (At this case water density)
g = Gravitational acceleration
Therefore the apparent weight will be 0.1526N
Explanation:
760 mmHg (millimeters of mercury) is a measure of atmospheric pressure. It represents the height of a column of mercury at which the static pressure at the bottom is equal to the atmospheric pressure.
1 atm = 760 mmHg = 101,325 Pa = 14.7 psi
