Respuesta :
Answer:
Rate of change of mass is given by,
[tex]\frac{dm}{dt}=18.816\,g/s[/tex]
Explanation:
In the question,
We have the Base Area of the vertical container = 14 cm x 16 cm
Now,
Let us take the height of the container = h
Rate of change of height with time, dh/dt = 0.21 cm/s = 2.1 mm/s
So,
Volume of the container = Base Area x Height
So,
V = 14 x 16 x h
V = 140 x 160 x h (because, 1 cm = 10 mm)
V = (22400)h
Now,
Volume of one of the candy = 50 mm³
Mass of the candy = 0.0200 g
So,
Density of the candy = Mass/ Volume
So,
[tex]Density=\frac{0.0200}{50}\\Density=0.0004[/tex]
Now,
V = (22400)h
On differentiating with respect to time, t, we get,
[tex]\frac{dV}{dt}=\frac{22400h}{dt}\\\frac{d}{dt}(\frac{mass}{density})=22400.\frac{dh}{dt}\\\frac{1}{density}.\frac{dm}{dt}=22400.\frac{dh}{dt}\\[/tex]
Therefore, on putting the value of density in the equation and also the value of rate of change of height with time, we get,
[tex]\frac{1}{density}.\frac{dm}{dt}=22400.\frac{dh}{dt}\\\frac{1}{0.0004}.\frac{dm}{dt}=22400(2.1)\\\frac{dm}{dt}=18.816\,g/s[/tex]
Therefore, the rate of change of mass with respect to time is given by,
[tex]\frac{dm}{dt}=18.816\,g/s[/tex]
