Answer:
technically both are correct but tech b is more correct because generally the older the vehicle the worse abs is newer vehicles have abs that no matter how hard you push on the brake peddle the wheels shouldnt lock up
Explanation:
Answer:

Explanation:
Given that:
The height of a triangular stabilizing fin on its stern is 1 ft tall
and it length is 2 ft long.
Temperature = 60 °F
The objective is to determine the drag on the fin when the submarine is traveling at a speed of 2.5 ft/s.
From these information given; we can have a diagrammatic representation describing how the triangular stabilizing fin looks like as we resolve them into horizontal and vertical component.
The diagram can be found in the attached file below.
If we recall ,we know that;
Kinematic viscosity v = 
the density of water ρ = 62.36 lb /ft³



which is less than < 5.0 × 10⁵
Now; For laminar flow; the drag on the fin when the submarine is traveling at 2.5 ft/s can be determined by using the expression:

where;
= strip area

Therefore;


Let note that y = 0.5x from what we have in the diagram,
so , x = y/0.5
By applying the rule of integration on both sides, we have:


Let U = (2-2y)
-2dy = du
dy = -du/2


![F_D = -0.568 [ \dfrac{\frac{1}{2}U^{ \frac{1}{2}+1 } }{\frac{1}{2}+1}]^0__2](https://tex.z-dn.net/?f=F_D%20%3D%20-0.568%20%5B%20%5Cdfrac%7B%5Cfrac%7B1%7D%7B2%7DU%5E%7B%20%5Cfrac%7B1%7D%7B2%7D%2B1%20%7D%20%20%7D%7B%5Cfrac%7B1%7D%7B2%7D%2B1%7D%5D%5E0__2)
![F_D = -0.568 [ \dfrac{2}{3}U^{\frac{3}{2} } ] ^0__2](https://tex.z-dn.net/?f=F_D%20%3D%20-0.568%20%5B%20%5Cdfrac%7B2%7D%7B3%7DU%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%20%20%5D%20%5E0__2)
![F_D = -0.568 [0 - \dfrac{2}{3}(2)^{\frac{3}{2} } ]](https://tex.z-dn.net/?f=F_D%20%3D%20-0.568%20%5B0%20-%20%20%5Cdfrac%7B2%7D%7B3%7D%282%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%20%20%5D)
![F_D = -0.568 [- \dfrac{2}{3} (2.828427125)} ]](https://tex.z-dn.net/?f=F_D%20%3D%20-0.568%20%5B-%20%5Cdfrac%7B2%7D%7B3%7D%20%282.828427125%29%7D%20%20%20%5D)


Answer:
L= 312.75 mm
Explanation:
given data
elastic modulus E = 106 GPa
cross-sectional diameter d = 3.9 mm
tensile load F = 1660 N
maximum allowable elongation ΔL = 0.41 mm
to find out
maximum length of the specimen before deformation
solution
we will apply here allowable elongation equation that is express as
ΔL =
....................1
put here value and we get L
L = 
solve it we get
L = 0.312752 m
L= 312.75 mm
Answer:
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