Answer:
- stress equation :
- Shear stress equation :
- cross sectional area of a beam equation : b*d
- cross sectional area of a shaft equation :
- shear stress at an angle to the axis of the member equation:
sin∅cos∅. - Normal stress at an angle to the axis of the member equation:
∅ - factor of safety equation :
- strain under axial loading equation:
Explanation:
The description of all the pieces to the equations
- stress equation : p = axial force, A = cross sectional area
- Shear stress equation : Q = calculated statistical moment, I = moment of inertia, v = calculated shear, b = width of beam
- cross sectional area of a beam equation : b*d b=width of beam, d =depth of beam
- cross sectional area of a shaft equation : d = shaft diameter
- shear stress at an angle to the axis of the member equation: sin∅cos∅. P = axial force, A = cross sectional area ∅ = given angle
- Normal stress at an angle to the axis of the member equation: ∅ p = axial force , A = cross sectional area, ∅ = given angle
- factor of safety equation :
- strain under axial loading equation: P = axial force, L = length, A = cross sectional area, E = young's modulus
Answer:
-273.16 °C
-459.677 °F
0 °K
0 °R
Explanation:
The lowest temperature is the absolute zero.
Absolute zero is at 0 degrees Kelvin, or 0 degrees Rankine, because these are absolute scales that have their zero precisely at the absolute zero.
Celsius and Fahrenheit degrees are relative scales, these have their zeroes above the absolute zero.
Celsius scale has the same degree separation as the Kelvin scale, but the zero is separated by 273.16 degrees. Therefore the lowest temperature in the Celsius scale is -273.16 °C.
The Fahrenheit degrees have the same degree separation as the Rankine degrees, and the zero is 459.67 degrees. Therefore the lowest temperature in the Fahrenheit scale is -459.67 °F.
Probably low heat and a longer time. Better sad than sorry
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:
Air mass sensors is the right answer i think
Explanation:
