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yawa3891 [41]
3 years ago
6

A tall, open container is full of glycerine. At what depth h below the surface of the glycerine is the pressure 2370 Pa greater

than atmospheric pressure? The density of glycerine is 1.26X10^3 kg/m^3
Physics
1 answer:
Mice21 [21]3 years ago
5 0

Answer:

So, at the depth of 24 cm below the surface of the glycerine the pressure is  2970 Pa. Hence, this is the required solution.

Explanation:

Given that,

Pressure exerted by the surface of glycerine, P = 2970 Pa and it is greater than atmospheric pressure.

The density of glycerine,  

We need to find the depth h below the surface of the glycerine. The pressure due to some depth is given by :

h = 0.24 meters

or

h = 24 cm

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What substances are acidic
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A substance with LOW pH. The lower the pH, the higher the acidity level is. pH has a direct effect on how acidic a substance is. For an example, battery acid. Battery acid has a pH level of about 1. This means that battery acid is very acidic. A substance is concidered acidic when it dips below the neutral pH level of 7.
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Earth is 149.6 million meters from the Sun and takes 365 days to make one complete revolution around the Sun. Mars is 227.9 mill
luda_lava [24]

Answer:

\frac{a_{r,earth}}{a_{r,mars}} = 2.325

Explanation:

The distance of Earth from the Sun is 149.6\times 10^{9}\,m and of Mars from the Sun is 227.9\times 10^{9}\,m. Let assume that both planets have circular orbits. The centripetal accelaration can be found by using the following expression:

a_{r} = \frac{v^{2}}{R}

Since planet has translation at constant speed, this formula is applied to compute corresponding speeds:

v=\frac{2\pi\cdot r}{\Delta t}

Earth:

v_{earth} = \frac{2\pi\cdot (149.6\times 10^{9}\,m)}{(365\,days)\cdot(\frac{24\,hours}{1\,day} )\cdot(\frac{3600\,s}{1\,h} )}

v_{earth}=29806.079\,\frac{m}{s}

Mars:

v_{mars} = \frac{2\pi\cdot (227.9\times 10^{9}\,m)}{(687\,days)\cdot(\frac{24\,hours}{1\,day} )\cdot(\frac{3600\,s}{1\,h} )}

v_{mars}=24124.244\,\frac{m}{s}

Now, centripetal accelarations can be found:

Earth:

a_{r,earth} = \frac{(29806.079\,\frac{m}{s} )^{2}}{149.6\times 10^{9}\,m}

a_{r,earth} = 5.939\times 10^{-3}\,\frac{m}{s^{2}}

Mars:

a_{r,mars} = \frac{(24124.244\,\frac{m}{s} )^{2}}{227.9\times 10^{9}\,m}

a_{r,mars} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}

The ratio of Earth's centripetal acceleration to Mars's centripetal acceleration is:

\frac{a_{r,earth}}{a_{r,mars}} = \frac{5.939}{2.554}

\frac{a_{r,earth}}{a_{r,mars}} = 2.325

7 0
3 years ago
Describe an imaginary process that satisfies the second law but violates the first law of thermodynamics.
N76 [4]

Answer:

Explanation:

First last of thermodynamics, just discusses the changes that a system is undergoing and the processes involved in it. It explains conservation of energy for a system undergoing changes or processes.

Second law of thermodynamics helps in defining the process and also the direction of the processes. It tells about the possibility of a process or the restriction of a process. It states that the entropy of a system always increases.

For this to occur the energy contained by a body has to diminish without converting to work or internal energy. So imagine a machine which works with less than efficiency, this means there are losses but they don’t show up anywhere. But the energy is obtained from a higher energy source to lower.

The easy way to do this is with an imaginary device that extracts zero-point energy to heat a quantity of gas. Energy is being created, so the first law is violated, and the entropy of the system is increasing as the gas heats up.

First law is violated since the energy conversion don't apply but the direction of work is applied so second law is satisfied.

7 0
3 years ago
A horizontal air diffuser operates with inlet velocity and specific enthalpy of 250 m/s and 270.11 kj/kg, repectively, and exit
Ivahew [28]

Answer: c) 90 m/s

Explanation:

Given

Invest velocity, v1 = 250 m/s

Inlet specific enthalpy, h1 = 270.11 kJ/kg = 270110 J/kg

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Outlet velocity, v2 = ?

0 = Q(cv) - W(cv) + m[(h1 - h2) + 1/2(v1² - v2²) + g(z1 - z2)]

0 = Q(cv) + m[(h1 - h2) + 1/2(v1² - v2²)]

0 = [(h1 - h2) + 1/2(v1² - v2²)]

Substituting the values of the above, we get

0 = [(270110 - 297310) + 1/2 ( 250² - v²)

0 = [-27200 + 1/2 (62500 - v²)]

27200 = 1/2 (62500 - v²)

54400 = 62500 - v²

v² = 62500 - 54400

v² = 8100

v = √8100

v = 90 m/s

4 0
3 years ago
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