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## Questions

### Subject:Calculus

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Question:

A spherical balloon is being inflated at the rate of 10 cu in/sec. Find the rate of change of the area when the balloon has a radius of 6 in.

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\$(V = 4 /3 π r ^3\$) and \$( A = 4 π r ^2\$) \$(d V /d t = 3*4 /3 π r ^2 dr/dt\$) \$( dAdt=2*4πrdr/dt\$) the value of dV/dt is given in the question so \$(dV/d t = 10 in^3 / s e c\$) If we substitute the value into the volume equation we can find dr/dt like so \$(10 = 4 /3 π 3 r ^2 d r/ d t\$) then set r = 6 we get \$(d r/ d t = 10 /452.39 = .0221\$) Then move on to solve this equation for dA/dt. substituting dr/dt value from other equation and setting r = 6 again ; \$(d A /d t = 10 / 3 = 3.33 i n ^2 / s e c\$)

### Subject:Chemical Engineering

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Question:

A container holds a mixture of three nonreacting gases: n1 moles of the first gas with molar specific heat at constant volume C1, and so on. Find the molar specific heat at constant volume of the mixture, in terms of the molar specific heats and quantitites of the three separate gases.

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Concept:- Heat capacity C of a body as the ratio of the amount of heat energy Q transferred to a body in any process to its corresponding temperature change ΔT. C = Q/ΔT So, Q = C ΔT Each species will experience the equal temperature change. If the gas has n molecules, then Q will be, Q = nC ΔT Solution:- If the gas has n1 moles, then the amount of heat energy Q1 transferred to a body having heat capacity C1 will be, Q1 = n1C1 ΔT Similarly, if the gas has n2 moles, then the amount of heat energy Q2 transferred to a body having heat capacity C2 will be, Q2 = n2C2 ΔT And if the gas has n3 moles, then the amount of heat energy Q3 transferred to a body having heat capacity C3 will be, Q3 = n3C3 ΔT As, each species will experience the same temperature change, thus, Q = Q1 + Q2 + Q3 = n1C1 ΔT + n2C2 ΔT + n3C3 ΔT Dividing both the sides by n = n1 + n2 + n3 and ΔT, then we will get, Q/nΔT = (n1C1 ΔT + n2C2 ΔT + n3C3 ΔT)/ nΔT As, Q/nΔT = C, thus, C=ΔT (n1C1 + n2C2 + n3C3)/ nΔT = (n1C1 + n2C2 + n3C3)/ n = n1C1 + n2C2 + n3C3/ n1 + n2 + n3 From the above observation we conclude that, the molar specific heat at constant volume of the mixture would be n1C1 + n2C2 + n3C3/ n1 + n2 + n3.

### Subject:Algebra

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Question:

Solve the equation 5(-3x - 2) - (x - 3) = -4(4x + 5) + 13

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Given the equation 5(-3x - 2) - (x - 3) = -4(4x + 5) + 13 Multiply factors. -15x - 10 - x + 3 = -16x - 20 +13 Group like terms. -16x - 7 = -16x - 7 Add 16x + 7 to both sides and write the equation as follows 0 = 0 The above statement is true for all values of x and therefore all real numbers are solutions to the given equation.

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