Answer:
Explanation:
efficiency = (energy output / energy input)
energy input = ivt where i is current = 6 A, v is volts = 140 and t time = 11 seconds
energy input = 6 × 140 × 11 = 9240 J
energy output = 0.64 × 9240 = 5913.6 J
510 pound-force = 2268.593 N
at height h, potential energy = mgh = 5913.6 J
2268.593 × h = 5913.6
h = 5913.6 / ( 2268.593) = 2.61 m
Answer:
yes it is possible
Explanation:
dislocation are if two type edge and screw dislocations
edge dislocation is a defect where an extra half plane is inside the lattice.
and screw dislocation is one in which can be assumed as the first half of the crystal slips over another.
These dislocation can coexist together where the line direction and burger vectors are neither parallel nor perpendicular then at that condition both dislocation screw and edge will coexist
Answer:
Explanation:
To obtain 240 V from 12 V batteries they must be connected in series. The number needed is ...
240/12 = 20 . . . batteries needed
__
The current draw will be ...
(3000 W)/(240 V) = 12.5 A
Then the time available from the battery stack is ...
(120 Ah)/(12.5 A) = 9.6 h
The motor can run 9.6 hours from the series connection.
Answer:
The solution code is written in Python.
- def largest3(num1, num2, num3):
- largest = num1
- if(largest < num2):
- largest = num2
-
- if(largest < num3):
- largest = num3
-
- return largest
- first_num = int(input("Enter first number: "))
- second_num = int(input("Enter second number: "))
- third_num = int(input("Enter third number: "))
- largest_number = largest3(first_num, second_num, third_num)
- print("The largest number is " + str(largest_number))
Explanation:
<u>Create function largest3</u>
- Firstly, we can create a function <em>largest3 </em>which take 3 numbers (<em>num1, num2, num3</em>) as input. (Line 1).
- Please note Python uses keyword <em>def </em>to denote a function. The code from Line 2 - 10 are function body of <em>largest3</em>.
- Within the function body, create a variable,<em> largest</em>, to store the largest number. In the first beginning, just tentatively assign<em> num1 </em>to<em> largest</em>. (Line 2)
- Next, proceed to check if the current "<em>largest</em>" value smaller than the<em> num2 </em>(Line 4). If so, replace the original value of largest variable with <em>num2</em> (Line 5).
- Repeat the similar comparison procedure to<em> </em><em>num3</em> (Line 7-8)
- At the end, return the final value of "<em>largest</em>" as output
<u>Get User Input</u>
- Prompt use input for three numbers (Line 13 -15) using Python built-in <em>input</em> function.
- Please note the input parts of codes is done outside of the function <em>largest3</em>.
<u>Call function to get largest number and display</u>
- We can simply call the function<em> largest </em>by writing the function name <em>largest</em> and passing the three user input into the parenthesis as arguments. (Line 17)
- The function <em>largest </em>will operate on the three arguments and return the output to the variable <em>largest_number</em>.
- Lastly, print the output using Python built-in <em>print</em> function. (Line 18)
- Please note the output parts of codes is also done outside of the function<em> largest3</em>.
Answer:
Explanation:
Initial conditions


Final conditions


Steady flow energy equation
![\dot{m}\left [ h_1+\frac{v_1^2}{2}+gz_1\right ]+\dot{Q}=\dot{m}\left [ h_2+[tex]\frac{v_2^2}{2}+gz_2\right ]+\dot{W}](https://tex.z-dn.net/?f=%5Cdot%7Bm%7D%5Cleft%20%5B%20h_1%2B%5Cfrac%7Bv_1%5E2%7D%7B2%7D%2Bgz_1%5Cright%20%5D%2B%5Cdot%7BQ%7D%3D%5Cdot%7Bm%7D%5Cleft%20%5B%20h_2%2B%5Btex%5D%5Cfrac%7Bv_2%5E2%7D%7B2%7D%2Bgz_2%5Cright%20%5D%2B%5Cdot%7BW%7D)
![\dot{m}\left [ c_pT_1+\frac{0^2}{2}+g0\right ]+\dot{Q}=\dot{m}\left [ c_pT_2+\frac{0^2}{2}+g0\right ]+\dot{W}](https://tex.z-dn.net/?f=%5Cdot%7Bm%7D%5Cleft%20%5B%20c_pT_1%2B%5Cfrac%7B0%5E2%7D%7B2%7D%2Bg0%5Cright%20%5D%2B%5Cdot%7BQ%7D%3D%5Cdot%7Bm%7D%5Cleft%20%5B%20c_pT_2%2B%5Cfrac%7B0%5E2%7D%7B2%7D%2Bg0%5Cright%20%5D%2B%5Cdot%7BW%7D)
![\dot{m}c_p\left [ T_1-T_2\right ]+\left [ -5hp\right ]=\dot{W} -5\times 746\times 3.4121](https://tex.z-dn.net/?f=%5Cdot%7Bm%7Dc_p%5Cleft%20%5B%20T_1-T_2%5Cright%20%5D%2B%5Cleft%20%5B%20-5hp%5Cright%20%5D%3D%5Cdot%7BW%7D%20-5%5Ctimes%20746%5Ctimes%203.4121)
![-4\dot{m}-\dot{m}\times 0.24\times \left [ 400-60\right ]](https://tex.z-dn.net/?f=-4%5Cdot%7Bm%7D-%5Cdot%7Bm%7D%5Ctimes%200.24%5Ctimes%20%5Cleft%20%5B%20400-60%5Cright%20%5D)


