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# Tutor profile: Daniel A.

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Daniel A.
Founder and a mechanical engineer with economics minor
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## Questions

### Subject:Mechanical Engineering

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Question:

Explain the different methods of heat transfer between objects?

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Daniel A.

There are three ways heat is transferred. The first is through convection heat transfer, which deals with heat transfer through a fluid (in mechanical engineering this defined as a liquid or gas). The fluid becomes the medium in which heat is transferred. Examples of this can be, blowing into a soup and air from a heater. The second one is conduction heating which is heat transfer through solid objects. So your stovetop is a great example of this. It uses highly conductive (of heat) material to transfer it into your pot or whatever you're heating. Another exmaple is you touching a metal staircase railing that is cold to the touch. That is conduction heating as well, since it was transfered by contact. The third form of heat transfer is through radiation. For example, the sun is a great example of this. Heat is transferred to us through radiation. Note that there is a specturm of raditaiton transmitted from the sun and some of it falls under this form of heat transfer as well as others falling under light and other radiations. The last thing to mention is that all three forms of heat transfer are at works all the time. For example let's take the example of a home during winter. Inside the home you will have a heater that will force hot air through the whole house using convection heating. At the same time, that same heated air is transferring to the outside air using the the walls using conduction heat transfer. Also, the outside cold air is transferring into the home through convection of the air and conduciton through the wall. And the sun from the outside is also transferring heat to the house and surrounding area using radiation heat transfer. Another point worth mentioning is that heat transfer shouldn't only be associated with really hot objects or fluids. It is the transfer of thermal energy. Even if something feels cold to the touch, doesn't mean it doesn't have any thermal energy. Hope this clears things up!

### Subject:Calculus

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Question:

Find the derivative of 34(x^4)+(y/2)-28=0

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Daniel A.

Now let's break this down into sections. 1st section: 34(x^4) d/dx = 34(4)*(x^(4-1)) d/dx = (136)*(x^(3)) d/dx = 136(x^3) In here we followed the basic derivative rule of bringing the power down to multiply with the coefficient (which is 34 in this case) and subtracting one from the power (which makes it from 4 to 3) Now let's follow the same rule for the second part: y/2 d/dx = y/2 d/dx = (1/2)*(y^1) Note, this simplification makes it easier to understand the derivative process. Any variable or number that doesn't show a power is always to the power of 1. The reason we don't write it is because it is redundant. For example, 2^1 is still 2, but writing just 2 make things easier and cleaner. It has been broken up in this section to clearly show you what happens when you differentiate a formula. d/dx = (1/2)*1*(y^(1-1)) d/dx = (1/2)(y^(0)) *Note, anything to the power of 0 is always equal to 1 d/dx = (1/2)*(1) d/dx = 1/2 The third section is just a number (-28). This doesn't have any variable attached to it, so the differentiation is a simple process. -28 d/dx = -28 d/dx = 0 Any number or constant derived gives you an answer of 0. So now let's combine our answers: 136(x^3)+(1/2)+0 = 0 136(x^3)+(1/2) = 0 And that is the final answer.

### Subject:Algebra

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Question:

Solve for x 3(x^2)+98 = (y/2) 23y-2x=17

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Daniel A.

Now since we have two variables, x & y, and we have two formulas given we can solve for one of them. First step is to isolate y. I'd choose the first formula since is is much easier to isolate y. Here are the steps: 3(x^2) + 98 = (y/2) 2*(3(x^2) + 98) = ((y/2))*2) *Note, this follows the rule that whatever you do to one side, you have to do to the other. 6(x^2) + 196 = y *In here we multiplied the 2 to the content inside the bracket, which gives you 6, 196 and leaves y by itself. Second step is to plug in the new answer for y into the second formula so we can solve for x. 23y-2x=17 23(6(x^2) + 196) = 17 138(x^2) + 4508 = 17 So far so good, we've plugged in the new value of y and have simplified the formula by multiplying into the brackets. 138(x^2) + 4508 -4508 = 17 - 4508 138(x^2) = -4491 At this point we subtracted 4508 from both sides to isolate x on its own. (138(x^2))/138 =( -4491)/138 x^2 = -4491/138 I'm keeping it as a fraction for simplicity's sake. Now, we get to the tricky part. In order to get x by itself, we need to find the square root of it. However, on the other side of the equation we have a negative number. Depending on your level of algebra there are two answers to this questions from this point on. Intermediate level: squareroot(x^2) = squareroot(-4491/138) x = undefined answer aka it doesn't exist. Advanced Level: squareroot(x^2) = squareroot(-4491/138) squareroot(x^2) = squareroot(-1 *(4491/138)) x = i*squareroot(4491/138) x =5.705i This becomes an introdution of the imaginary number i, which is a value that when squared gives you a negative number and can be the square root of a negative number as well.

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