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MAVERICK [17]
3 years ago
12

Steam is leaving a 4-L pressure cooker whose operating pressure is 150 kPa. It is observed that the amount of liquid in the cook

er has decreased by 0.6 L in 40 min after the steady operating conditions are established, and the cross-sectional area of theexit opening is 8mm2. Determine
a)The mass flow rate of the steam and the exit velocity
b)The total and flow energies of the steam per unit mass
c)The rate at which energy leaves the cooker by steam.

Physics
1 answer:
jeka57 [31]3 years ago
3 0

Answer: a) Mr = 2.4×10^-4kg/s

V = 34.42m/a

b) E = 173J

Ø = 2693.1J

c) Er = 0.64J/s

Explanation: Please find the attached file for the solution

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The specific weight of sea water is 10.1 kN/m^3. Convert to lbs/in^3.
Viktor [21]

Answer:

0.03719 lbs/in³

Explanation:

Specific weight is given by multiplying the density of an object to the acceleration due to gravity.

\gamma =\rho g\\\Rightarrow \rho=\frac{gamma}{g}\\\Rightarrow \rho=\frac{10.1\times 10^3}{9.81}\\\Rightarrow \rho=1029.562\ kg/m^3

1\ kg=2.20462\ lb

1\ m=39.3701\ in

\\\Rightarrow 1029.562\ kg/m^3=\frac{1029.562\times 2.20462}{39.3701^3}=0.03719\ lbs/in^3

So,

10.1\ kN/m^3=0.03719\ lbs/in^3

8 0
2 years ago
A spherical balloon has a radius of 7.40 m and is filled with helium. Part A How large a cargo can it lift, assuming that the sk
malfutka [58]

Answer:

The mass of the cargo is M  =  188.43 \ kg

Explanation:

From the question we are told that

    The radius of the spherical balloon is  r =  7.40 \ m

     The mass of the balloon is  m = 990\ kg  

The volume of the spherical balloon is mathematically represented as

     V  =  \frac{4}{3} * \pi r^3

substituting values

      V  =  \frac{4}{3} * 3.142 *(7.40)^3

      V  =  1697.6 \ m^3

The total mass  the balloon can lift is mathematically represented as

     m =  V (\rho_h - \rho_a)

where \rho_h is the density of helium with a  value of

       \rho_h  =  0.179 \ kg /m^3

and  \rho_a is the density of air with a value of

        \rho_ a  = 1.29 \ kg / m^3

substituting values

          m =  1697.6 ( 1.29  - 0.179)

         m =  1886.0  \ kg

Now the mass of the cargo is mathematically evaluated as

        M  =  1886.0 - 1697.6

        M  =  188.43 \ kg

       

5 0
3 years ago
Scientific evidence documents the pattern of evolution. The evidence exists in a variety of categories. What are these categorie
Sauron [17]

Answer:

Explanation:

Scientific evidence takes note of the pattern of evolution. The evidence exists in a variety of categories, including direct observation of evolutionary change, the fossil record, homology, and biogeography.

Examples of the categories with their respective examples are:

Direct observation of evolutionary change: Development of drug resistant bacteria

Fossil record: Discovery of transitional forms of horses, Discovery of shells of extinct species

Homology: Similarities in mammalian forelimbs, Same genetic code in fireflies and tobacco plants, Vestigial pelvis in right whales

Biogeography: Similarity of endemic island species to nearby mainland species, The high concentration of marsupial species in Australia

3 0
3 years ago
What is shown in the figure above
Firdavs [7]

A single magnetic field is shown.

4 0
2 years ago
A real object is 10.0 cm to the left of a thin, diverging lens having a focal length of magnitude 16.0 cm. What is the location
amm1812

Answer:

A)6.15 cm to the left of the lens

Explanation:

We can solve the problem by using the lens equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where

q is the distance of the image from the lens

f is the focal length

p is the distance of the object from the lens

In this problem, we have

f=-16.0 cm (the focal length is negative for a diverging lens)

p=10.0 cm is the distance of the object from the lens

Solvign the equation for q, we find

\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}

q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm

And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is

A)6.15 cm to the left of the lens

6 0
3 years ago
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