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Vitek1552 [10]
3 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]3 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:

\mathbf{h_2 =1021.9 \  mm}

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At temperature T_2 = 70^0 C  the density of the water is \rho = 977.8 \ kg/m^3

The mass of the water is  \rho V = \rho _1 V_1 = \rho _2 V_2

Thus; we can say \rho _1 V_1 = \rho _2 V_2;

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The area of the water before heating is:

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A_1 = \dfrac{1000000}{1000}

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A_2 = 1000.5 \ mm^2

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h_2 = \dfrac{V_2}{A_2}

h_2 = \dfrac{1022400}{1000.5}

\mathbf{h_2 =1021.9 \  mm}

Hence the depth of the heated hot  water is \mathbf{h_2 =1021.9 \  mm}

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