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Burka [1]
3 years ago
4

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.
A) Find the pressures pa and pb at surfaces a and b in the tube, respectively. Use patmos to denote atmospheric pressure.
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?
C) Assume p atmos=1.00atm. What is the gas pressure pgas?
Physics
1 answer:
Natasha_Volkova [10]3 years ago
8 0

ANSWER:

PART A) Find the pressures pa and pb at surfaces a and b in the tube, respectively. Use patmos to denote atmospheric pressure.

Express answers, separated by a comma,in terms on one or both of the variables

            Pgas

Pa,Pb =P atmos

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?

It is useul to draw a horizontal line because it allows you to relate pressure in the mercury at crticial points in the tube. For example, even though you are not given the pressure at point c, you know that Pc=Patmos, because point C is at the same height as any point on surface B. At the same time, you can also use the hydrostatic pressure equation to relate the pressure at point C to that surfarce A, where pressure is determined by the gas pressure in the box.

Part c)Assume p atmos=1.00atm. What is the gas pressure pgas?

Conversion factor between atmospheres and pascals

1atm= 101,300 Pa

Find the height of the liquid column above point C

What is the height hc of the of the mercury column above point C?

hc= h1- h2

Pgas= Patmos - Pmg(h1 - h2) Pa

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Answer:

he formula for the gravitational force includes the gravitational constant, which has a value . The unit of the gravitational force is Newtons (N). Fg = gravitational force between two objects ( ) G = gravitational constant ( ) m1 = mass of the first object (kg)

Explanation:

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4 years ago
A gas is compressed from 600cm3 to 200cm3 at a constant pressure of 450kPa . At the same time, 100J of heat energy is transferre
Paraphin [41]

Answer: 80J

Explanation:

According to the first principle of thermodynamics:  

<em>"Energy is not created, nor destroyed, but it is conserved."  </em>

Then this priciple (also called Law) relates the work and the transferred heat exchanged in a system through the internal energy U, which is neither created nor destroyed, it is only transformed. So, in this especific case of the compressed gas:

\Delta U=Q+W  (1)

Where:

\Delta U is the variation in the internal (thermal) energy of the system (the value we want to find)

Q=-100J is the heat transferred out of the gas (that is why it is negative)

W is the work is done on the gas (as the gas is compressed, the work done on the gas must be considered positive )

On the other hand, the work done on the gas is given by:

W=-P \Delta V  (2)

Where:

P=450kPa=450(10)^{3}Pa is the constant pressure of the gas

\Delta V=V_{f}-V_{i} is the variation in volume of the gas

In this case the initial volume is V_{i}=600{cm}^{3}=600(10)^{-6}m^{3} and the final volume is V_{f}=200{cm}^{3}=200(10)^{-6}m^{3}.

This means:

\Delta V=200(10)^{-6}m^{3}-600(10)^{-6}m^{3}=-400(10)^{-6}m^{3}  (3)

Substituting (3) in (2):

W=-450(10)^{3}Pa(-400(10)^{-6}m^{3})  (4)

W=180J  (5)

Substituting (5) in (1):

\Delta U=-100J+180J  (6)

Finally:

\Delta U=80J  This is the change in thermal energy in the compression process.

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3 years ago
Find the moment of inertia about each of the following axes for a rod that is 0.360 {cm} in diameter and 1.70 {m} long, with a m
mamaluj [8]

The complete question is;

Find the moment of inertia about each of the following axes for a rod that is 0.36 cm in diameter and 1.70m long, with a mass of 5.00 × 10 ^(−2) kg.

A) About an axis perpendicular to the rod and passing through its center in kg.m²

B) About an axis perpendicular to the rod and passing through one end in kg.m²

C) About an axis along the length of the rod in kg.m²

Answer:

A) I = 0.012 kg.m²

B) I = 0.048 kg.m²

C) I = 8.1 × 10^(-8) kg.m²

Explanation:

We are given;

Diameter = 0.36 cm = 0.36 × 10^(−2) m

Length; L = 1.7m

Mass;m = 5 × 10^(−2) kg

A) For an axis perpendicular to the rod and passing through its center, the formula for the moment of inertia is;

I = mL²/12

I = (5 × 10^(−2) × 1.7²)/12

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B) For an axis perpendicular to the rod and passing through one end, the formula for the moment of inertia is;

I = mL²/3

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I = (5 × 10^(−2) × 1.7²)/3

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C) For an axis along the length of the rod, the formula for the moment of inertia is; I = mr²/2

We have diameter = 0.36 × 10^(−2) m, thus radius;r = (0.36 × 10^(−2))/2 = 0.18 × 10^(−2) m

I = (5 × 10^(−2) × (0.18 × 10^(−2))^2)/2

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