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ale4655 [162]
3 years ago
5

Which pulley system will allow the armchair to be lifted with the least amount of effort?

Chemistry
2 answers:
lyudmila [28]3 years ago
7 0

ANSWER:

I think it's C

Explanation:

I just learned about this in science class and I'm pretty sure you need 2 pulleys to make it easier to pull

guapka [62]3 years ago
5 0

Answer:

Option D.

Explanation:

Option D. Since tension experimented by the rope is equal to one sixth of the weight of armchair.

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A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce
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Explanation:

The given data is as follows.

       Volume of lake = 15 \times 10^{6} m^{3} = 15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}

        Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = 5.6 \times 15 \times 10^{9} mg

                                                                    = 84 \times 10^{9} mg

                                                                    = 84 \times 10^{3} kg

Flow rate of river is 50 m^{3} sec^{-1}

Volume of water in 1 day = 50 \times 10^{3} \times 86400 liter

                                          = 432 \times 10^{7} liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are 2.9792 \times 10^{10} mg or 2.9792 \times 10^{4} kg

Flow rate of sewage = 0.7 m^{3} sec^{-1}

Volume of sewage water in 1 day = 6048 \times 10^{4} liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = 1.8144 \times 10^{10} mg or 1.8144 \times 10^{4}kg

Therefore, total concentration of lake after 1 day = \frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l

                                        = 6.8078 mg/l

                 k_{D} = 0.2 per day

       L_{o} = 6.8078

Hence, L_{liquid} = L_{o}(1 - e^{-k_{D}t}

             L_{liquid} = 6.8078 (1 - e^{-0.2 \times 1})  

                             = 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

                                                             = 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

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what mass of chromium (III) sulfide is produced from the reaction of 13.2 g of sulfur with excess chromium?
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Answer: Cr2S3

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