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Vitek1552 [10]
4 years ago
11

A man weighs 145 lb on earth.Part ASpecify his mass in slugs.Express your answer to three significant figures and include the ap

propriate units.Part BSpecify his mass in kilograms.Express your answer to three significant figures and include the appropriate units.Part CSpecify his weight in newtons.Express your answer to three significant figures and include the appropriate units.Part DIf the man is on the moon, where the acceleration due to gravity is gm = 5.30 ft/s2, determine his weight in pounds.Express your answer to three significant figures and include the appropriate units.Part EDetermine his mass in kilograms.Express your answer to three significant figures and include the appropriate units.
Engineering
1 answer:
adell [148]4 years ago
3 0

Answer:

<em>a) 4.51 lbf-s^2/ft</em>

<em>b) 65.8 kg</em>

<em>c) 645 N</em>

<em>d) 23.8 lb</em>

<em>e) 65.8 kg</em>

<em></em>

Explanation:

Weight of the man on Earth = 145 lb

a) Mass in slug is...

32.174 pound = 1 slug

145 pound = x slug

x = 145/32.174 = <em>4.51 lbf-s^2/ft</em>

b) Mass in kg is...

2.205 pounds = 1 kg

145 pounds = x kg

x = 145/2.205 = <em>65.8 kg</em>

c) Weight in Newton = mg

where

m is mass in kg

g is acceleration due to gravity on Earth = 9.81 m/s^2

Weight in Newton = 65.8 x 9.81 = <em>645 N</em>

d) If on the moon with acceleration due to gravity of 5.30 ft/s^2,

1 m/s^2 = 3.2808 ft/s^2

x m/s^2 = 5.30 ft/s^2

x = 5.30/3.2808 = 1.6155 m/s^2

weight in Newton = mg = 65.8 x 1.6155 = 106

weight in pounds = 106/4.448 = <em>23.8 lb</em>

e) The mass of the man does not change on the moon. It will therefore have the same value as his mass here on Earth

mass on the moon = <em>65.8 kg</em>

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Answer:

W_{12}=-53.9056KJ

Part A:

Q=-7.03734 KJ/Kg (-ve sign shows heat is getting out)

Part B:

Q=1.5265KJ/Kg (Heat getting in)

The value of Q at constant specific heat is approximately 361% in difference with variable specific heat and at constant specific heat Q has opposite direction (going in) than Q which is calculated in Part B from table A-23. So taking constant specific heat is not a good idea and is questionable.

Explanation:

Assumptions:

  1. Gas is ideal
  2. System is closed system.
  3. K.E and P.E is neglected
  4. Process is polytropic

Since Process is polytropic so  W_{12} =\frac{P_{2}V_{2}-P_{1}V_{1}}{1-n}

Where n=1.25

Since Process is polytropic :

\frac{V_{2}}{V_{1}}=(\frac{P_{1}}{P_{2}})^{\frac{1}{1.25}} \\V_{2}= (\frac{P_{1}}{P_{2}})^{\frac{1}{1.25}} *V_{1}

V_{2}= (\frac{0.7}{11})^{\frac{1}{1.25}} *0.262\\V_{2}=0.028924 m^3

Now,W_{12} =\frac{P_{2}V_{2}-P_{1}V_{1}}{1-n}

W_{12} =\frac{11*0.028924-0.7*0.262}{1-1.25}(\frac{10^{5}N/m^2}{1 bar})(\frac{1  KJ}{10^{3}Nm})

W_{12}=-53.9056KJ

We will now calculate mass (m) and Temperature T_2.

m=\frac{P_{1}V_{1}}{RT_{1}}\\ m=\frac{0.7*0.262}{\frac{8.314KJ}{44.01Kg.K}*320}(\frac{10^{5}N/m^2}{1 bar})(\frac{1  KJ}{10^{3}Nm})\\m=0.30338Kg

T_{2} =\frac{P_{2}V_{2}}{Rm}\\ m=\frac{11*0.028924}{\frac{8.314KJ}{44.01Kg.K}*0.30338}(\frac{10^{5}N/m^2}{1 bar})(\frac{1  KJ}{10^{3}Nm})\\T_{2} =555.14K

Part A:

According to energy balance::

Q=mc_{v}(T_{2}-T_{1})+W_{12}

From A-20, C_v for Carbon dioxide at 300 K is 0.657 KJ/Kg.k

Q=0.30338*0.657(555.14-320)+(-53.9056)

Q=-7.03734 KJ/Kg (-ve sign shows heat is getting out)

Part B:

From Table A-23:

u_{1} at 320K = 7526 KJ/Kg

u_{2} at 555.14K = 15567.292 (By interpolation)

Q=m(\frac{u(T_{2})-u(T_{1})}{M} )+W_{12}

Q=0.30338(\frac{15567.292-7526}{44.01} )+(-53.9056)

Q=1.5265KJ/Kg (Heat getting in)

The value of Q at constant specific heat is approximately 361% in difference with variable specific heat and at constant specific heat Q has opposite direction (going in) than Q which is calculated in Part B from table A-23. So taking constant specific heat is not a good idea and is questionable.

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