Explanation:
The pressure exerted by a column of liquid of height h and density ρ is given by the hydrostatic pressure equation p = ρgh, where g is the gravitational acceleration
1. Life is meaningless.
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Answer:
To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.
The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.
Thus let the density of the material be Pm
50*3= 47.5*Pm
Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube
Thus with an average density of 2.8gram per centimeter cube
50*2.8= (50-x)*3.16
(50-x)= (50*2.8)/3.16
50-x=44.3
x=50-44.3= 5.7
Explanation:
To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.
The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.
Thus let the density of the material be Pm
50*3= 47.5*Pm
Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube
Thus with an average density of 2.8gram per centimeter cube
50*2.8= (50-x)*3.16
(50-x)= (50*2.8)/3.16
50-x=44.3
x=50-44.3= 5.7
I'm assuming we're applying the standard Integral form of the calculation of work. The solution is provided in the image.
Answer:
Volume, V = 13564.8 cubic feet
Explanation:
It is given that,
Radius of the cylindrical tank, r = 12 feet
Height of the tank, h = 30 feet
We need to find the water that can be held by a cylindrical tank i.e. we need to find the volume of the tank. It is given by :


V = 13564.8 cubic feet
So, the water held by the tank is 13564.8 cubic feet. Hence, this is the required solution.