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sattari [20]
3 years ago
15

Three forces act on a statue. Force F⃗ 1F→1 (magnitude 45.0 NN) points in the +x-direction, Force F⃗ 2F→2 (magnitude 105 NN) poi

nts in the +y-direction, and force F⃗ 3F→3 (magnitude 235 NN) is at an angle of 36.9∘∘ from the -x-direction and 53.1∘∘ from the +y-direction. counterclockwise from the +x-direction. Find the x- and y-components of the net external force F⃗ F→ on the statue
Physics
1 answer:
Trava [24]3 years ago
5 0

Answer:

Resultant force = (232.93î + 246.10j) N

x-component of the resultant force = (+232.93î) N

y-component of the resultant force = (+246.1j) N

Explanation:

The net external force on the statue is equal to the resultant force on the statue.

And the resuphant force is a vector sum of all the other forces acting on the statue.

Force 1 = (45î) N

Force 2 = (105j) N

Force 3 = (235cos 36.9°)î + (235 sin 36.9°)j = (187.93î + 141.10j) N

Resultant force = (Force 1) + (Force 2) + (Force 3)

Resultant force = 45î + 105j + (187.93î + 141.10j) = (232.93î + 246.10j) N

Hope this helps!!!

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What change in entropy occurs when a 0.15 kg ice cube at -18 °C is transformed into steam at 120 °c 4.
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<u>Answer:</u> The change in entropy of the given process is 1324.8 J/K

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The processes involved in the given problem are:

1.)H_2O(s)(-18^oC,255K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\3.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\4.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\5.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(120^oC,393K)

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\Delta S=m\times C_{p,m}\times \ln (\frac{T_2}{T_1})      .......(1)

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C_{p,m} = specific heat capacity of medium

m = mass of ice = 0.15 kg = 150 g    (Conversion factor: 1 kg = 1000 g)

T_2 = final temperature

T_1 = initial temperature

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\Delta S=m\times \frac{\Delta H_{f,v}}{T}      .......(2)

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\Delta S = Entropy change

m = mass of ice

\Delta H_{f,v} = enthalpy of fusion of vaporization

T = temperature of the system

Calculating the entropy change for each process:

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We are given:

m=150g\\C_{p,s}=2.06J/gK\\T_1=255K\\T_2=273K

Putting values in equation 1, we get:

\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K

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We are given:

m=150g\\\Delta H_{fusion}=334.16J/g\\T=273K

Putting values in equation 2, we get:

\Delta S_2=\frac{150g\times 334.16J/g}{273K}\\\\\Delta S_2=183.6J/K

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We are given:

m=150g\\C_{p,l}=4.184J/gK\\T_1=273K\\T_2=373K

Putting values in equation 1, we get:

\Delta S_3=150g\times 4.184J/g.K\times \ln(\frac{373K}{273K})\\\\\Delta S_3=195.9J/K

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We are given:

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\Delta S_2=\frac{150g\times 2259J/g}{373K}\\\\\Delta S_2=908.4J/K

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We are given:

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Putting values in equation 1, we get:

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Total entropy change for the process = [21.1+183.6+195.9+908.4+15.8]J/K=1324.8J/K

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