To develop this problem it is necessary to apply the concepts performed to the absolute pressure based on the reference pressure (atmospheric) and the pressure that is generated due to the height of the column of the measured liquid.
In mathematical terms the previous concept can be expressed as

Where
Atmospheric Pressure
Density
g = Gravitational acceleration
h = Height
Our values are given as

g = 9.8m/s


Replacing we have then that



Therefore the absolute pressure in the test section is 99.9019kPa
I'm pretty sure it's the neutron star.
Answer:
Most of the frogs have webbed feet which helps them swim. The thin skin between the toes helps them to push through the water. The frog's feet is webbed ,So they can easily swim.. The skin between toes are flexible soo they can freely push and pull it
Explanation:
Answer:
a)α= 53.13°
b)The velocity at the highest point = 15 m/s
The acceleration at the highest point = 9.8
c)h=15 m
V=18.02 m/s
Explanation:
Speed of water ,u= 25 m/s
So the horizontal component of speed u = u cos α
Given that horizontal distance cover by water in 3 s is 45 m.
So We know that in projectile motion horizontal acceleration is zero.
In horizontal direction
Distance = Velocity x time
45 = u cos α x 3
u cos α = 45
45 = 25 cos α x 3
cos α = 45/75
α= 53.13°
So the velocity at the highest point = u cos α
The velocity at the highest point = 15 m/s
The acceleration at the highest point = 9.8
Now the velocity along vertical direction(Vo) = u sin α
Vo= 25 sin 53.13°
Vo =20 m/s


h=15 m
So at 15 m above the ground water will strike .
The y-component of velocity after 3 sec
Vy= Vo - g t
Vy = 20 - 10 x 3
Vy= -10 m/s
The horizontal component of velocity will remain 15 m/s.
The resultant velocity

V=18.02 m/s