Answer:
point_dist = math.sqrt((math.pow(x2 - x1, 2) + math.pow(y2 - y1, 2))
Explanation:
The distance formula is the difference of the x coordinates squared, plus the difference of the y coordinates squared, all square rooted. For the general case, it appears you simply need to change how you have written the code.
point_dist = math.sqrt((math.pow(x2 - x1, 2) + math.pow(y2 - y1, 2))
Note, by moving the 2 inside of the pow function, you have provided the second argument that it is requesting.
You were close with your initial attempt, you just had a parenthesis after x1 and y1 when you should not have.
Cheers.
Answer:
The solution for the given problem is done below.
Explanation:
M1 = 2.0
= 0.3636
= 0.5289
= 0.7934
Isentropic Flow Chart: M1 = 2.0 , 
= 1.8
T1 = 
(1.8)(288K) = 653.4 K.
In order to choke the flow at the exit (M2=1), the above T0* must be stagnation temperature at the exit.
At the inlet,
T02= 
= (1.8)(288K) = 518.4 K.
Q= Cp(T02-T01) = 
= 135.7*
J/Kg.
Given data:
•) applied voltage = 15 V
•). Resistance = 1000 ohm
Required:
•). The magnitude of current= ?
•••••••••••••SOLUTION•••••••••••••
We can find the relation ship between current, voltage and resistance with the help of Ohms law.
According to ohms law;
V= IR.
Rearranging the above equation;
I= V/ R
Putt the values in the above equation; we get
I= 15V/ 1000ohm
I = 0.015 A( ampere)
••••••••••••••• CONCLUSION•••••••
The value of the current would be 0.15 ampere when Resistance is equal to 1000 and that of Voltage is equal to 15 V.
Answer:
minimum flow rate provided by pump is 0.02513 m^3/s
Explanation:
Given data:
Exit velocity of nozzle = 20m/s
Exit diameter = 40 mm
We know that flow rate Q is given as
where A is Area
minimum flow rate provided by pump is 0.02513 m^3/s
Answer:
A phrase from: who loves life
Explanation:
