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Sauron [17]
3 years ago
8

Complex machines are defined by

Engineering
1 answer:
Lera25 [3.4K]3 years ago
6 0

Answer:

their multiple moving parts.

Explanation:

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Which is equal to a temperature of 50°F?<br><br> 18°C<br> 46°C<br> 10°C<br> 32°C
Ludmilka [50]

Answer:

10°C degrees po ang sagot

8 0
2 years ago
Read 2 more answers
Gold forms a substitutional solid solution with silver. Compute the number of gold atoms per cubic centimeter for a silver-gold
evablogger [386]

Answer:

Compute the number of gold atoms per cubic centimeter = 9.052 x 10^21 atoms/cm3

Explanation:

The step by step and appropriate substitution is as shown in the attachment.

From number of moles = Concentration x volume

number of moles = number of particles/ Avogadro's number

Volume = mass/density, the appropriate derivation to get the number of moles of atoms

5 0
3 years ago
Suppose within your web browser you click on a link to obtain a web page. The IP address for the associated URL is already cache
aleksandrvk [35]

Answer:

All the detailed steps are mentioned in pictures.

Explanation:

See attached pictures.

8 0
3 years ago
A reversible process and an irreversible process both have the same________ between the same two states. a. Internal energy b. W
Vlad1618 [11]

Answer:

a) Internal energy

Explanation:

As we know that internal energy is a point function so it did not depends on the path ,it depends  at the initial and final states of process.All point function property did not depends on the path.Internal energy is a exact function.

Work and heat is a path function so these depend on the path.They have different values for different path between two states.Work and heat are in exact function.

We know that in ir-reversible process entropy will increase so entropy will be different for reversible and ir-reversible processes.

5 0
3 years ago
The gas expanding in the combustion space of a reciprocating engine has an initial pressure of 5 MPa and an initial temperature
Anit [1.1K]

Answer:

a). Work transfer = 527.2 kJ

b). Heat Transfer = 197.7 kJ

Explanation:

Given:

P_{1} = 5 Mpa

T_{1} = 1623°C

                       = 1896 K

V_{1} = 0.05 m^{3}

Also given \frac{V_{2}}{V_{1}} = 20

Therefore, V_{2} = 1  m^{3}

R = 0.27 kJ / kg-K

C_{V} = 0.8 kJ / kg-K

Also given : P_{1}V_{1}^{1.25}=C

   Therefore, P_{1}V_{1}^{1.25} = P_{2}V_{2}^{1.25}

                     5\times 0.05^{1.25}=P_{2}\times 1^{1.25}

                     P_{2} = 0.1182 MPa

a). Work transfer, δW = \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}

                                  \left [\frac{5\times 0.05-0.1182\times 1}{1.25-1}  \right ]\times 10^{6}

                              = 527200 J

                             = 527.200 kJ

b). From 1st law of thermodynamics,

Heat transfer, δQ = ΔU+δW

   = \frac{mR(T_{2}-T_{1})}{\gamma -1}+ \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}

  =\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W

  =\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200

  = 197.7 kJ

6 0
3 years ago
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