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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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cheryl has a mug that she says is made up of matter. heather says that the hot chocolate inside the cup is made up of matter, to
9966 [12]
Well,  basically both Cheryl and Keaton is right.

matter is described as Physical substance that possessed space and mass.

But the thing is, one matter is very small in size and cannot be seen with naked eyes. But everything around us are made by matters

hope this helps


3 0
3 years ago
A harmonic wave is traveling along a rope. It is observed that the oscillator that generates the wave completes 37.6 vibrations
tangare [24]

Answer:

\lambda = 25.79\ cm

Explanation:

given,

Wave vibrates = 37.6

time = 27.9 s

maximum distance travel = 450 cm

time = 11.3 s

wavelength = ?

frequency of wave

f=\dfrac{37.6}{27.9}

f = 1.35 Hz

Speed of wave

v = \dfrac{450}{11.3}

v = 39.82 cm/s

wavelength of wave

v = fλ

\lambda =\dfrac{v}{f}

\lambda =\dfrac{34.82}{1.35}

\lambda = 25.79\ cm

Hence, wavelength of the wave is equal to 25.79 cm.

8 0
3 years ago
4. Using the bone density of 2.0 kg/m3, calculate the mass of an adult femur bone that has a volume of 0.00027 m3.
kolbaska11 [484]

Answer:

\boxed{\sf Mass \ of \ an \ adult \ femur \ bone = 0.00054 \ kg}

Given:

Bone density = 2.0 kg/m³

Volume of bone (V) = 0.00027 m³

To Find:

Mass of an adult femur bone (m).

Explanation:

\sf \implies Density = \frac{Mass (m)}{Volume (V)} \\ \\ \sf \implies \frac{Mass}{Volume} = Density \\ \\ \sf \implies Mass = Density \times Volume \\ \\ \sf \implies Mass = 2.0 \ kg/ \cancel{m^{3}} \times 0.00027 \ \cancel{m^{3}} \\ \\ \sf \implies Mass = 0.00054 \ kg

6 0
3 years ago
Explain why an egg thrown at a concrete wall will break, but an egg thrown at a sheet hanging from the ceiling will not. (Be sur
Firlakuza [10]

An egg thrown at a concrete wall will break, but an egg thrown at a sheet hanging from the ceiling will not due to high momentum and acceleration.

<h3>Why an egg thrown at a concrete wall will break?</h3>

An egg thrown at a concrete wall will break, but an egg thrown at a sheet hanging from the ceiling will not because the momentum and acceleration increases when the egg is thrown downward due to gravity but when we throw an egg in the vertical direction, they move against gravity so the momentum and acceleration decreases.

So we can conclude that an egg thrown at a concrete wall will break, but an egg thrown at a sheet hanging from the ceiling will not due to high momentum and acceleration.

Learn more about momentum here: brainly.com/question/7538238

#SPJ1

4 0
2 years ago
Calculate the wavelength of each frequency of electromagnetic radiation: a. 100.2 MHz (typical frequency for FM radio broadcasti
Natalka [10]

Answer:

a). 100.2 MHz (typical frequency for FM radio broadcasting)

The wavelength of a frequency of 100.2 Mhz is 2.99m.

b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)

The wavelength of a frequency of 1070 khz is 280.3 m.

c. 835.6 MHz (common frequency used for cell phone communication)

The wavelength of a frequency of 835.6 Mhz is 0.35m.

Explanation:

The wavelength can be determined by the following equation:

c = \lambda \cdot \nu  (1)

Where c is the speed of light, \lambda is the wavelength and \nu is the frequency.  

Notice that since it is electromagnetic radiation, equation 1 can be used. Remember that light propagates in the form of an electromagnetic wave.

<em>a). 100.2 MHz (typical frequency for FM radio broadcasting)</em>

Then, \lambda can be isolated from equation 1:

\lambda = \frac{c}{\nu} (2)

since the value of c is 3x10^{8}m/s. It is necessary to express the frequency in units of hertz.

\nu = 100.2 MHz . \frac{1x10^{6}Hz}{1MHz} ⇒ 100200000Hz

But 1Hz = s^{-1}

\nu = 100200000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{100200000s^{-1}}

\lambda = 2.99 m

Hence, the wavelength of a frequency of 100.2 Mhz is 2.99m.

<em>b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)</em>

<em> </em>

\nu = 1070kHz . \frac{1000Hz}{1kHz} ⇒ 1070000Hz

But  1Hz = s^{-1}

\nu = 1070000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{1070000s^{-1}}

\lambda = 280.3 m

Hence, the wavelength of a frequency of 1070 khz is 280.3 m.

<em>c. 835.6 MHz (common frequency used for cell phone communication) </em>

\nu = 835.6MHz . \frac{1x10^{6}Hz}{1MHz} ⇒ 835600000Hz

But  1Hz = s^{-1}

\nu = 835600000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{835600000s^{-1}}

\lambda = 0.35 m

Hence, the wavelength of a frequency of 835.6 Mhz is 0.35m.

6 0
3 years ago
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