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den301095 [7]
3 years ago
13

Why does teachers grade things that are not due yet​

Engineering
2 answers:
pav-90 [236]3 years ago
4 0
I think because if you’ve already turned it in they might as well grade asap instead of waiting
White raven [17]3 years ago
4 0
I’m here for the points
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ra1l [238]
I need more details to your question
4 0
2 years ago
Read 2 more answers
1: asha started abusness with 30.000
svetoff [14.1K]

Answer:

Explanation:adrive with visual acutity of 20/30 can just decipher asing adistance 20ft from asing determine the maximum destance from the sing which drivers with the flowing visual acuities will able to see the same sing 20/15 20/50

4 0
2 years ago
Koch traded Machine 1 for Machine 2 when the fair market value of both machines was $60,000. Koch originally purchased Machine 1
Mariana [72]

Answer:

Koch's adjusted basis in machine 2 after the exchange is $60,000

Explanation:

given data

fair market value = $60,000

originally purchased Machine 1 = $76,900

Machine 1 adjusted basis = $40,950

Machine 2 seller purchase = $64,050

Machine 2 adjusted basis = $55,950

solution

As he exchanged machine for another at $60,000

and this exchanged in fair market

so adjusted basis =  $50,000

Adjusted basis is the price of the item that affects the factors that are considered price. These factors usually include taxes, depreciation value, and other costs of acquiring and maintaining a given item. Adjusted basis is important so the right amount to sell

Adjusted basis increases when a person deducts expenses from factor taxes and operating statements

so Koch's adjusted basis in machine 2 after the exchange is $60,000

3 0
3 years ago
What would you like to visit friends?
serious [3.7K]
If i would visit one of my friends it’s c
8 0
3 years ago
A one-dimensional plane wall of thickness 2L=80 mm experiences uniform thermal generation of q= 1000 W/m^3 and is convectively c
Eduardwww [97]

Answer:

h=1.99998\ W/m^2.C

k=33.333\ W/m.C

Explanation:

Considering the one dimensional and steady state:

From Heat Conduction equation considering the above assumption:

\frac{\partial^2T}{\partial x^2}+\frac{\dot e_{gen}}{k}=0       Eq (1)

Where:

k is thermal Conductivity

\dot e_{gen} is uniform thermal generation

T(x) = a(L^2-x^2)+b

\frac{\partial\ T(x)}{\partial x}=\frac{\partial\ a(L^2-x^2)+b}{\partial x}=-2ax\\\frac{\partial^2\ T(x)}{\partial x^2}=\frac{\partial^2\ -2ax}{\partial x^2}=-2a

Putt in Eq (1):

-2a+\frac{\dot e_{gen}}{k}=0\\ k=\frac{\dot e_{gen}}{2a}\\ k=\frac{1000}{2*15}\\ k=33.333\ W/m.C

Energy balance is given by:

Q_{convection}=Q_{conduction}

h(T_L-T_{inf})=-k(\frac{dT}{dx}) _L     Eq  (2)

T(x) = a(L^2-x^2)+b

Putting x=L

T(L) = a(L^2-L^2)+b\\T(L)=b\\T(L)=40^oC

\frac{dT}{dx}=\frac{d(a(L^2-x^2)+b}{dx}=-2ax\\Put\ x\ =\ L\\\frac{dT}{dx}=-2aL\\(\frac{dT}{dx})_L=-2*15*0.04=-1.2

From Eq (2)

h=\frac{-k*-1.2}{(40-20)} \\h=\frac{-33.333*-1.2}{(40-20)}\\h=1.99998\ W/m^2.C

7 0
3 years ago
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