Physics, asked by XxRishabhRathorexX, 22 hours ago

A vertical container with base area measuring 14.0 cm by 17.0 cm is being filled with identical pieces of candy, each with a volume of 50.0 mm3 and a mass of 0.0200 g. Assume that the volume of the empty spaces between the candies is negligible. If the height of the candies in the container increases at the rate of 0.250 cm/s, at what rate (kilograms per minute) does the mass of the candies in the container increase?​

Answers

Answered by sabanashareef
2

Explanation:

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Question

A vertical container with base area measuring 14.0 cm by 17.0 cm is being filled, with identical pieces of candy, each with a volume of 50.0 mm3 and a mass of 0.0200 g. Assume that the volume of empty spaces between the candies in negligible. If the height of the candies in the container increases at the rate of 0.250 cm/s, at what rate (kilograms per minute) does the mass of the candies in the container increase?

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Solution

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The mass density of the candy is

ρ=Vm=50.0 mm30.0200 g=4.00×10−4 g/mm3=4.00×10−4kg/cm3.

If we neglect the volume of the empty spaces between the candies, then the total mass of the candies in the containers when filled to height h is M=ρAh, where A=(14.0 cm)(17.0 cm)=238 cm2 is the base of the container that remains unchanged. Thus, the rate of mass change is given by

dtdM=dtd(ρAh)=ρAdtdh=(4.00×10−4kg/cm3)(238 cm2)(0.250 cm/s)

=0.0238 kg/s=1.43 kg/min

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Answered by priya150377
1

Answer:

1.43 kg/min is the correct answer.

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