Answer:
Explanation:
Given
Diver is at a depth of 
inside the water
As we go down pressure increases due to the weight of liquid column above us
Pressure difference is given by 
Where 
=density of liquid
h=depth
g=acceleration due to gravity
So external-internal pressure difference is
Answer
It increases to B, then it decreases.
When the stone thrown upwards, it moves vertically upwards to a maximum height then is starts to come down. Due to gravitational force, the velocity of the stone projected upwards decreases but when coming down wards is increases.
The trajectory shown in the diagrams represents an example of an object projected upwards.
We can then says that the vertical component increases from A to B, then it decreases to C.
The type
of composition that igneous rocks have that contains mostly plagioclase
feldspar and about 30 percent dark silicate minerals would be andesitic. I am hoping that this
answer has satisfied your query about and it will be able to help you.
While the boy is sitting on the chair it creates a force downward on the chair and therefore the chair takes it and gives off the equal amount of force. So while he is putting force downward the chair is putting the same force upward.
Complete question is;
a. Two equal sized and shaped spheres are dropped from a tall building. Sphere 1 is hollow and has a mass of 1.0 kg. Sphere 2 is filled with lead and has a mass of 9.0 kg. If the terminal speed of Sphere 1 is 6.0 m/s, the terminal speed of Sphere 2 will be?
b. The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1. The masses remain 1.0 kg and 9.0 kg, The terminal speed (in m/s) of Sphere 2 will now be
Answer:
A) V_t = 18 m/s
B) V_t = 10.39 m/s
Explanation:
Formula for terminal speed is given by;
V_t = √(2mg/(DρA))
Where;
m is mass
g is acceleration due to gravity
D is drag coefficient
ρ is density
A is Area of object
A) Now, for sphere 1,we have;
m = 1 kg
V_t = 6 m/s
g = 9.81 m/s²
Now, making D the subject, we have;
D = 2mg/((V_t)²ρA))
D = (2 × 1 × 9.81)/(6² × ρA)
D = 0.545/(ρA)
For sphere 2, we have mass = 9 kg
Thus;
V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρA))]
V_t = 18 m/s
B) We are told that The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1.
Thus;
Area of sphere 2 = 3A
Thus;
V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρ × 3A))]
V_t = 10.39 m/s
