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Mekhanik [1.2K]
2 years ago
15

Hello :) how to do 25 (b) ?

Physics
1 answer:
sp2606 [1]2 years ago
3 0

Answer:

Explanation:

You first solve point a by finding the components < R_x; R_y > of the resulting vector, which will be

< 0 + 40 \times cos 30\° + 30 \times cos (- 50\°); 20 + 40 \times sin 30\° + 30 \times sin (-50\°) >

Notice that since the 50° angle is taken clockwise, we consider it negative, while the 30° is considered positive.

Plugging the values in a calculator or a spreadsheet (if you are using excel or similar programs, don't forget to convert the angle in radians!) you will get the components, namely < 53.92;17.02 >

At this point magnitude is defined as R = \sqrt{R_x^2 + R_y^2} while for direction you first take the ratio R_y/R_x and again, with a calculator, you get the inverse tangent of it. The result will be a number between -\frac{\pi}2 and \frac\pi2 (-90° and 90°) which tells you the angle the line containing your vector forms with the positive x axis, in our case it's tan^-^1 (53.92/17.02) \approx 17.52 (ie, it sits on the red line)

At this point, based on the fact that both are positive, the end point of the vector is in the first quadrant.

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James Joule (after whom the unit of energy is named) claimed that the water at the bottom of Niagara Falls should be warmer than
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0.12 K

Explanation:

height, h = 51 m

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According to the transformation of energy

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m x g x h = m x c x ΔT

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Calculate the specific heat at constant volume of water vapor, assuming the nonlinear triatomic molecule has three translational
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I) c=1385.667\frac{J}{kg K}

II)The difference from the value obtained on part I is: 2000-1385.67 =614.33 \frac{J}{Kg K}

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Explanation:

From the problem we have the molar mass given M=18\frac{gr}{mol} of water vapor and at constant volume condition. It's important to say that the vapour molecules have 3 transitionsl and 3 rotational degrees of freedom and the rotational motion no contribution.

Part I

Calculate the specific heat at constant volume of water vapor, assuming the nonlinear triatomic molecule has three translational and three rotational degrees of freedom and that vibrational motion does not contribute. The molar mass of water is 18.0 g/mol=0.018kg/mol.

Let C_v (\frac{J}{Kg K}) the molar heat capacity at constant volume and this amount represent the quantity of heat absorbed by mole.

Let C (\frac{J}{Kg K}) the specific heat capcity this value represent the heat capacity aboserbed by mass.

For the problem we have a total of 6 degrees of freedom and from the thoery we know that for each degree of freedom the molar heat capacity at constant volume is given by C_v =\frac{R}{2} so the total for the 6 degrees of freedom would be:

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If we replace all the values we have:

c=\frac{24.942\frac{J}{mol K}}{0.018\frac{kg}{mol}}=1385.667\frac{J}{kg K}

So on this case the specific heat capacity with constant volume and with three translational and three rotational degrees of freedom is c=1385.667\frac{J}{kg K}

Part II

The actual specific heat of water vapor at low pressures is about 2000 J/(kg * K). Compare this with your calculation.

The difference from the value obtained on part I is: 2000-1385.67 =614.33 \frac{J}{Kg K}

The possible reason of this difference is that the vibrational motion can increase the value, since if we take in count this factor we will have a higher heat capacity, because molecules with vibrational motion require more heat to vibrate and necessary higher specific heat capacity.

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