To solve the problem we will first start considering the Pressure given the hydrostatic definition of the product between the density, the gravity and the depth. We will define the area where the liquid acts and later we will use the definition of the force as a product between the pressure and the area to calculate the force given in the two depths. The gauge pressure at the depth x will be
This pressure acts on the strip of area
The force acting on that strip is given by,
To evaluate the force, we will then consider the integral of the pressure as a function of the Area, or the integral of the previously found terms.
Evaluating at the initial depth of 1.8m and the final depth of 4.4 we have then that,
Therefore the Net force will be
At terminal velocity, drag force becomes equal to weight. Therefore:
weight = bv²
0.0023 x 9.81 = b x 9.1²
b = 2.72 x 10⁻⁴
Answer:
KINETIC ENERGY=K.E=0.00245 Joules ≈ 0.003 Joules
Explanation:
mass=m=0.1g=0.0001kg
velocity=v=7.0m/s
kinetic energy=K.E.=?
as we know that
hope it will help ^_^
Answer:
stabilize
I think this should be the answer...
17.3 L.
<h3>Explanation</h3>
The object appears 
lighter in water than in the air. Water has supplied that 170 N of buoyant force.
The size of the buoyant force on an object in water is the same as the weight of water that the object has displaced. The buoyant force on the metal object here is 170 N. The object must have displaced water of the same weight.
.
Mass of water displaced:
.
Volume of water displaced:
The density of water at room temperature is 
. Each kilogram of water will occupy a volume of 1 dm³ (one cubic decimeter), which is the same as 1 L (one liter).
.
Volume of the object:
The object is completely under water. As a result, the volume of the object will be the same as the volume of water displaced. The volume of the object is also 17.3 L.
