Answer:b
Explanation:It’s only applies to the left-turning traffic
Answer:
a
Explanation:
the others dont make sense to be an engineer
Answer:
The displacement from t = 0 to t = 10 s, is -880 m
Distance is 912 m
Explanation:
. . . . . . . . . . A
integrate above equation we get
from information given in the question we have
t = 1 s, s = -10 m
so distance s will be
-10 = 12 - 1 + C,
C = -21
we know that acceleration is given as
[FROM EQUATION A]
Acceleration at t = 4 s, a(4) = -24 m/s^2
for the displacement from t = 0 to t = 10 s,
the distance the particle travels during this time period:
let v = 0,
t = 2 s
Distance ![= [s(2) - s(0)] + [s(2) - s(10)] = [1\times 2 - 2^3] + [(12\times 2 - 2^3) - (12\times 10 - 10^3)] = 912 m](https://tex.z-dn.net/?f=%3D%20%5Bs%282%29%20-%20s%280%29%5D%20%2B%20%5Bs%282%29%20-%20s%2810%29%5D%20%3D%20%5B1%5Ctimes%202%20-%202%5E3%5D%20%2B%20%5B%2812%5Ctimes%202%20-%202%5E3%29%20-%20%2812%5Ctimes%2010%20-%2010%5E3%29%5D%20%3D%20912%20m)
Answer:
I don't know ☺️☺️☺️❌‼️
Explanation:
I don't understand this question
Answer:
In the steel: 815 kPa
In the aluminum: 270 kPa
Explanation:
The steel pipe will have a section of:
A1 = π/4 * (D^2 - d^2)
A1 = π/4 * (0.8^2 - 0.7^2) = 0.1178 m^2
The aluminum core:
A2 = π/4 * d^2
A2 = π/4 * 0.7^2 = 0.3848 m^2
The parts will have a certain stiffness:
k = E * A/l
We don't know their length, so we can consider this as stiffness per unit of length
k = E * A
For the steel pipe:
E = 210 GPa (for steel)
k1 = 210*10^9 * 0.1178 = 2.47*10^10 N
For the aluminum:
E = 70 GPa
k2 = 70*10^9 * 0.3848 = 2.69*10^10 N
Hooke's law:
Δd = f / k
Since we are using stiffness per unit of length we use stretching per unit of length:
ε = f / k
When the force is distributed between both materials will stretch the same length:
f = f1 + f2
f1 / k1 = f2/ k2
Replacing:
f1 = f - f2
(f - f2) / k1 = f2 / k2
f/k1 - f2/k1 = f2/k2
f/k1 = f2 * (1/k2 + 1/k1)
f2 = (f/k1) / (1/k2 + 1/k1)
f2 = (200000/2.47*10^10) / (1/2.69*10^10 + 1/2.47*10^10) = 104000 N = 104 KN
f1 = 200 - 104 = 96 kN
Then we calculate the stresses:
σ1 = f1/A1 = 96000 / 0.1178 = 815000 Pa = 815 kPa
σ2 = f2/A2 = 104000 / 0.3848 = 270000 Pa = 270 kPa
