There are NO true statements on that list of choices.
The pressure increase at the bottom of the pool after they enter the pool and float is 106.103 Pa.
<h3>What is absolute pressure?</h3>
Absolute pressure is the force that exists in a space when there is no matter present, or when there is a perfect vacuum. This absolute zero serves as the baseline for measurements in absolute pressure. The measurement of barometric pressure is the greatest illustration of an absolute referenced pressure. In order to determine absolute pressure, a complete vacuum is used. In contrast, gauge pressure is the amount of pressure that is measured in relation to atmospheric pressure, also referred to as barometric pressure.
given,
diameter = 6 m
depth = h = 1.5 m
Atmospheric pressure = P₀ = 10⁵ Pa
a) absolute pressure
P = P₀ + ρ g h
P = 10⁵ + 1000 x 10 x 1.5
P = 1.15 x 10⁵ Pa
b) When two person enters into the pool,
mass of the two person = 150 Kg
weight of water level displaced exists equal to the weight of person.




Area of pool 


Height of the water rise



P = ρ g h
P = 1000 x 10 x 0.0106
P = 106.103 Pa
To learn more about absolute pressure refer to:
brainly.com/question/17200230
#SPJ4
The work done will be equal to the potential energy of the piano at the final position
P.E=m.g.h
.consider the plank the hypotenuse of the right triangle formed with the ground
.let x be the angle with the ground=31.6°
.h be the side opposite to the angle x (h is the final height of the piano)
.let L be the length of the plank
sinx=opposite side / hypotenuse
= h/L
then h=L.sinx=3.49×sin31.6°=0.638m
weight w=m.g
m=w/g=3858/10=385.8kg
Consider Gravity g=10m/s2
then P.E.=m.g.h=385.8kg×10×0.638=2461.404J
then Work W=P.E.=2451.404J
Answer:
Because you hit the break?
To solve this problem we will use the linear motion description kinematic equations. We will proceed to analyze the general case by which the analysis is taken for the second car and the tenth. So we have to:

Where,
x= Displacement
= Initial velocity
a = Acceleration
t = time
Since there is no initial velocity, the same equation can be transformed in terms of length and time as:

For the second cart

When the tenth car is aligned the length will be 9 times the initial therefore:

When the tenth car has passed the length will be 10 times the initial therefore:

The difference in time taken from the second car to pass it is 5 seconds, therefore:

From the first equation replacing it in the second one we will have that the relationship of the two times is equivalent to:


From the relationship when the car has passed and the time difference we will have to:



Replacing the value found in the equation given for the second car equation we have to:

Finally we will have the time when the cars are aligned is


The time when you have passed it would be:


The difference between the two times would be:

Therefore the correct answer is C.