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Marianna [84]
3 years ago
9

To practice tactics box 13.1 hydrostatics. in problems about liquids in hydrostatic equilibrium, you often need to find the pres

sure at some point in the liquid. this tactics box outlines a set of rules for thinking about such hydrostatic problems.
part a) find the pressures pa and pb at surfaces a and b in the tube, respectively. use patmos to denote atmospheric pressure.

express your answers, separated by a comma, in terms of one or both of the variables pgas and patmos.

part b) as stated in rule 3 in the tactics box, it is always convenient to use horizontal lines in hydrostatic problems. in each one of the following sketches, a different horizontal line is considered. which sketch would be more useful in solving the problem of finding the gas pressure?
Physics
1 answer:
OLEGan [10]3 years ago
5 0

Answer:

A. The pressure denoted as Pa and Pb at the surfaces of A and B in the tube is

PA= Pgas

PB= Patmos

B. The second sketch

C. The gas pressure is

Pgas= Patmos+ rho.g(h2-h1)

= 1atm + rho.g (h2-h1)

Explanation:

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Answer: 52 kg skydiver: 9.09 m/s and 522.55 s

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For skydivers, when falling through the air, the forces acting on it are gravitational and drag forces. At a certain point, drag force equals gravitational force, which is constant on any part of the planet, producing a net force that is zero. Since there is no net force, there is no acceleration and, consequently, velocity is constant. When that happens, the person reached the <u>Terminal</u> <u>Velocity</u>.

Drag Force and Velocity are proportional to the squared speed. So, terminal velocity is given by:

F_{G}=F_{D}

mg=\frac{1}{2}C \rho Av_{T}^{2}

v_{T}=\sqrt{\frac{2mg}{\rho CA} }

where

m is mass in kg

g is acceleration due to gravitational force in m/s²

ρ is density of the fluid in kg/m³

C is drag coefficient

A is area of the object in the fluid in m²

Calculating:

The 52kg skydiver has terminal velocity of:

v_{T}=\sqrt{\frac{2(52)(9.8)}{(1.21)(0.7)(0.14)} }

v_{T}= 9.09

The 95kg skydiver's terminal velocity is

v_{T}=\sqrt{\frac{2(95)(9.8)}{(1.21)(0.7)(0.14)} }

v_{T}= 12.3

The 52 kg and 95kg skydivers' terminal velocity are 9.09m/s and 12.3m/s, respectively.

The time each one will reach the floor will be:

52 kg at 9.09 m/s:

t=\frac{4750}{9.09}

t = 522.5

95 kg at 12.3 m/s:

t=\frac{4750}{12.3}

t = 386.2

The 52 kg and 95kg skydivers' time to reach the floor are 522.5 s and 386.2 s, respectively.

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