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N76 [4]
3 years ago
12

You come across an open container which is filled with two liquids. Since the two liquids have different density there is a dist

inct separation between them. Water fills the lower portion of the container to a depth of 0.209 m which has a density of 1.00 × 103 kg/m3. The fluid which is floating on top of the water is 0.300 m deep. If the absolute pressure on the bottom of the container is 1.049 × 105 Pa, what is the density of the unknown fluid? The acceleration due to gravity is g = 9.81 m/s2 and atmospheric pressure is P0 = 1.013 × 105 Pa.
Physics
1 answer:
Slav-nsk [51]3 years ago
4 0

Answer:

526.57 Pa

Explanation:

P ( pressure at the bottom of the container) = 1.049 × 10^5 pa

Using the formula of pressure in an open liquid

Pw ( pressure due to water) = ρhg where ρ is density of water in kg/m³, h is the height in meters, and g is acceleration due to gravity in m/s²

Pw = 1000 × 9.81 ×0.209 = 2050.29 Pa

P( atmospheric pressure) = 1.013 × 10^5 Pa

Pl ( pressure due to the liquid) = ρ(density of the liquid) × h (depth of the liquid) × g

Subtract each of the pressure from the absolute pressure at the bottom

P(bottom) - atmospheric pressure

(1.049 × 10^5) - (1.013 × 10^5) = 0.036 × 10^5 = 3600 Pa

subtract pressure due to water from the remainder

3600 - 2050.29 = 1549.71 Pa

1549.71 =  ρ(density of the liquid) × h (depth of the liquid) × g

ρ (density of the liquid) = 1549.71 / (h × g) = 1549.71 / (0.3 × 9.81) =526.57 Pa

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What is the reverse of vaporization?
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That is called condensation.

When the water vaporizes the liquid transforms into vapor which goes to the atmosphere. When the water vapor of the atmosphere condensates liquid water is formed. You can see condensation when you have a glass with cold water and drops of water form in the exterior of the glass: those drops are liquid water that formed when the vapor of the air that surrounds the glass cools due to the lower temperature of the surface of the glass.


8 0
3 years ago
Nerve cells transmit electric signals through their long tubular axons. These signals propagate due to a sudden rush of Na+ ions
iren [92.7K]

Answer:

I = \frac{9.12x10^{-8} C}{0.007 s}= 1.30285 x10^{-5} \frac{C}{s}=1.30285 x10^{-5} A = 13.02 \mu A

Explanation:

For this case we have the following info given:

Number of Na+ ions 5.7 x10^{11} ions

Each ion have a charge of +e and the crage of the electron is 1.6 x10^{-19}C

The time is given t = 7 ms if we convert this into seconds we got:

t = 7ms * \frac{1s}{1000 ms}= 0.007s

Now we can use the following formula given from the current passing thourhg a meter of nerve axon given by:

Q = Ne

Where N represent the number of ions, e the charge of the electron and Q the total charge

If we replace on this case we have this:

Q= 5.7x10^{11} * (1.6 x10^{-19}C) = 9.12x10^{-8} C

And from the general definition of current we know that:

I =\frac{Q}{t}

And since we know the total charge Q and the time we can replace:

I = \frac{9.12x10^{-8} C}{0.007 s}= 1.30285 x10^{-5} \frac{C}{s}=1.30285 x10^{-5} A = 13.02 \mu A

The current during the inflow charge in the meter axon for this case is 13.02 \mu A

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5 0
3 years ago
A projector is placed on the ground 22 ft. away from a projector screen. A 5.2 ft. tall person is walking toward the screen at a
Stella [2.4K]

Answer:

y = 67.6 feet,   y = 114.4/ (22 - 3t)

Explanation:

For this exercise let's use that light travels in a straight line and some trigonometric relationships, the symbols are in the attached diagram

Large triangle Projector up to the screen

         tan θ = y / L

For the small triangle. Projector up to the person

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The angle is the same, so we equate the two equations

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The distance from the screen (d), we look for it with kinematics

         v = d / t

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we replace

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         y = 114.4 (22 - 3t)⁻¹

This is the equation of the shadow height change as a function of time

For the suggested distance the shadow has a height of

           y = 114.4 / (22-13)

           y = 67.6 feet

7 0
3 years ago
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