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timofeeve [1]
3 years ago
13

Everyone's favorite flying sport disk can be approximated as the combination of a thin outer hoop and a uniform disk, both of di

ameter Dd=0.273 m. The mass of the hoop part is mh=0.120 kg and the mass of the disk part is md=0.050 kg. You would like to make a boomerang that has the same total moment of inertia, around its center, as the sport disk. The boomerang is to be constructed in the shape of an \"X,\" which can be approximated as two thin uniform rods joined at their midpoints. If the total mass of the boomerang is to be mb = 0.245 kg, what must be the length Lb of the boomerang?
Physics
1 answer:
Umnica [9.8K]3 years ago
5 0

Answer:

The length of the boomerang is 0.364 m

Explanation:

The moment of inertia is:

I=\frac{1}{2} m_{d} r^{2} +m_{h} r^{2}

Where

md = 0.05 kg

mh = 0.12 kg

r = d/2 = 0.273/2 = 0.1365 m

I=\frac{0.05*(0.1365)^{2} }{2} +(0.12)*(0.1365)^{2} =2.7x10^{-3} kgm^{2}

The length of the boomerang is:

L_{b} =\sqrt{\frac{12I}{m} } =\sqrt{\frac{12*2.7x10^{-3} }{0.245} } =0.364m

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If 2.40 g of KNO3 reacts with sufficient sulfur (S8) and carbon (C), how much P-V work will the gases do against an external pre
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-112.876J

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w_{pv} = - P_{external}  \triangle Volume

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External pressure is given as  1.00atm , therefore the work solely depends on the change in volume and since the reactants are solids, none of the reactants contribute to the volume. Hence,  V_i = 0.

To find the volume of the products, we need to first find the amount of moles of the product made from  2.40_gKNO_3, using the molar mass of  KNO_3  which is 101.1032 g/mol  

2.40_gKNO_3 . {\frac{1molKNO_3}{101.1032_g}} = 0.0237molKNO_3

Now let us convert moles of  KNO_3  into moles of CO_2 and N_2  using the stoichiometric ratios from our balanced equation of the reaction.

0.0237molKNO_3 . {\frac{24molCO_2}{16molKNO_3}} = 0.0356molCO_2

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0.0356molCO_2 . {\frac{44.01_g}{1molCO_2}} = 1.567_gCO_2

0.01185molN_2 . {\frac{28.014_g}{1molN_2}} = 0.332_gN_2

We will now proceed to convert grams into volume using the density values provided.

1.567_gCO_2 . {\frac{1L}{1.830_g}} = 0.856LCO_2

0.332_gN_2 . {\frac{1L}{1.165_g}} = 0.285LN_2

Summing up the two volumes, we get the final volume

0.856L + 0.258L = 1.114L = V_f

Plugging everything into the w_{pv} equation, we get:

w_{pv} = -1atm(1.114L - 0L) = -1.114L.atm

Finally, let us convert L.atm into joules using the conversion rate of;

1L.atm = 101.325J\\-1.114L.atm. {\frac{101.325J}{1L.atm}} = -112.876J

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