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:
<u>the thickness required would be 12 inch HMA and granular base layer of 6 inches</u>
Explanation:
structural number = 4.5
stone base course material coefficient = 0.13
hma material layer coefficient = 0.40
drainage coefficient = 0.90
we will use layered analysis procedure to get thickness
D1 >= sN1/a1
when we cross multiply,
sN1 = a1D1 >=sN1
D2 >= -sN2-sN1/a2m2
sN2* + sN1* >= sN2
D3 >= sN3-(sN1*+sN2*)/a2m2
where a1,a2,a3 = layer coefficient
d1 d2 d3 = actual thickness
m2,m3 = coefficient of base
a1 = 0.4
a1 = 0.13
sN = 4.5
m2 = 0.9
D1 >= sN1/a1 = 4.5/0.4
= 11.25
thickness of surface = 12 inches
a1D1 = 0.4x12 = 4.8
we have value of sN2 = 5.5
(5.5 -4.8)/(0.13*0.9)
= 0.7/0.117
= 5.9829 inches
approximately 6 inches
so the pavement will have 12inch HMA surface and 6 inches granular base layer.
Answer:
The option that identifies why the bicycle cannot yet be created as a model in the scenario is;
Suzanne forgot to include the exact units of measurement that should be used
Explanation:
The design sketch turned over to the team that will work on the prototype by Suzanne should present all aspects of the design that will enable others working on the design and that make use of the sketch to have a clear understanding of what is required of them
Given that Suzanne has included the numbers that explain the relationship between the sketch and the real world object, the scale that shows the ratios and proportions of the sketch and the actual bicycle has been provided, however, given that the the machinist still need more information, the units of the measurement indicated in the drawing was not included, therefore, the actual dimensions and size that gives the length of the parts of the sketch and of the prototype to be made cannot be determined.
Answer:
Today, multiples can be created from the process. high-relief projects boldly from the background, and elements of high-relief may be in the round, unattached to the background.
Answer:
Tmax= 46.0 lb-in
Explanation:
Given:
- The diameter of the steel rod BC d1 = 0.25 in
- The diameter of the copper rod AB and CD d2 = 1 in
- Allowable shear stress of steel τ_s = 15ksi
- Allowable shear stress of copper τ_c = 12ksi
Find:
Find the torque T_max
Solution:
- The relation of allowable shear stress is given by:
τ = 16*T / pi*d^3
T = τ*pi*d^3 / 16
- Design Torque T for Copper rod:
T_c = τ_c*pi*d_c^3 / 16
T_c = 12*1000*pi*1^3 / 16
T_c = 2356.2 lb.in
- Design Torque T for Steel rod:
T_s = τ_s*pi*d_s^3 / 16
T_s = 15*1000*pi*0.25^3 / 16
T_s = 46.02 lb.in
- The design torque must conform to the allowable shear stress for both copper and steel. The maximum allowable would be:
T = min ( 2356.2 , 46.02 )
T = 46.02 lb-in
