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# Tutor profile: Prabath R.

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Prabath R.
Tutor for 4 years
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## Questions

### Subject:Linear Algebra

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Question:

The cake production company want to start packing the cakes after one-hour the cake is baked. The initial temperature of the baked cake is 60℃ and it got 15 minutes to reduce temperature to 50℃.Room temperature is 20℃ and they can do packing after it reaches to 30℃.Can they start the packing process as they desired?

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Prabath R.

Solution dT/dt=k(T-20) 1/(T-20) dT/dt=k Integrating with respect to time we get, ln⁡〖(T-20)=kt+C〗 T-20=e^(kt+c) T= e^c.e^kt+20 -------------- (1) When t = 0, t = 60 60=e^c+ e^c=40 So that, T=40.e^kt+20 When temperature reduces to 30℃ time should exceed 60 min, If it is they can start packing. When t = 15, T = 50 50=40 e^k(15) +20 e^15k= 30/40 e^k=〖( 3/4)〗^(1/15) Substituting into (1), T=50.e^kt+20 So that let's consider T = 30℃ 30 = 50.〖( 3/4)〗^(t/15)+20 〖( 3/4)〗^(t/15)= 1/5 〖ln( 3/4)〗^(t/15)= ln 1/5 t/15= (ln⁡(0.2))/(ln⁡(0.75)) t=83.92 Answer The 60 min is exceeded, so they can’t start packing.

### Subject:Mechanical Engineering

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Question:

Project description  All necessary logic and calculations  Finite element analysis (Forces, thermal)  Necessary engineering drawings  Possible prototype design  General literature parts

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Prabath R.

All the points will be included in the project report,

### Subject:Differential Equations

TutorMe
Question:

The cake production company want to start packing the cakes after one-hour the cake is baked. The initial temperature of the baked cake is 60℃ and it got 15 minutes to reduce temperature to 50℃.Room temperature is 20℃ and they can do packing after it reaches to 30℃.Can they start the packing process as they desired?

Inactive
Prabath R.

Solution dT/dt=k(T-20) 1/(T-20) dT/dt=k Integrating with respect to time we get, ln⁡〖(T-20)=kt+C〗 T-20=e^(kt+c) T= e^c.e^kt+20 -------------- (1) When t = 0, t = 60 60=e^c+ e^c=40 So that, T=40.e^kt+20 When temperature reduces to 30℃ time should exceed 60 min, If it is they can start packing. When t = 15, T = 50 50=40 e^k(15) +20 e^15k= 30/40 e^k=〖( 3/4)〗^(1/15) Substituting into (1), T=50.e^kt+20 So that let's consider T = 30℃ 30 = 50.〖( 3/4)〗^(t/15)+20 〖( 3/4)〗^(t/15)= 1/5 〖ln( 3/4)〗^(t/15)= ln 1/5 t/15= (ln⁡(0.2))/(ln⁡(0.75)) t=83.92 Answer The 60 min is exceeded, so they can’t start packing.

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