The magnitude of shear stress (kPa) on an element a lying at a distance of 20 mm from neutral axis. (6 pts) are given, p = 2kn e = 210gpa distance from constant cease = 20mm = 0.2m.
<h3>What is the cantilever formula?</h3>
The cantilever beam equations (deflection) w = load. l = member length. e = young's modulus. i = the beam's moment of inertia.
- theta= 2000 2*210*10^ 9 * (0.006)^ four 12 [(0.1)^ 2 -(0.08)^ 2 ] =0.1587 rad
- ø at q i.e. x = 100 - 20 = 80mm = 0.08m from loose cease.
- theta = p/(2ei) * (l ^ 2 - x ^ 2) phase of facet a. i = (a ^ four)/12 for square.
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Answer:
All the time and to see if someone is behind you.
Explanation:
After you check your side mirrors you should check your review mirror to see who is behind you.
Answer:
<u>Eo = A/[-nB/A]^(1/n-1) + B/[-nB/A]^(n/n-1)</u>
Explanation:
<u>Step 1.</u>
Taking derivative of the equation with respect to 'r' we get:
d/dr(EN) = - A/r² - nB/r^(n+1)
Setting this equation to zero:
<u>Step 2.</u>
Solving for r:
- A/r² - nB/r^(n+1) = 0
A/r² + nB/r^(n+1) = 0
Ar^(n+1) + nBr² = 0
Ar^(n+1) = - nBr²
[r^(n+1)]/r² = - nB/A
r^(n+1-2) = - nB/A
r^(n-1) = - nB/A
Taking power 1/(n-1) on both sides:
r = [-nB/A]^(1/n-1)
This is the value of ro:
ro = [-nB/A]^(1/n-1)
<u>Step 3.</u>
Substituting value of ro in eqn we get value of Eo
<u>Eo = A/[-nB/A]^(1/n-1) + B/[-nB/A]^(n/n-1)</u>
Question:
John read the first 114 pages of a novel, which was 3 pages less than ⅓ of the novel. Write an equation to determine the total number of pages (P)
Answer:
114 = ⅓P - 3
Explanation:
Given
Number of pages read = 114
Total pages in novel = p
The relationship between the pages read by John and the total pages is analysed as follows:
3 less than ⅓ of total pages means:
⅓ of total pages - 3
Recall that P represents the total pages in the novel
So, the expression becomes
⅓ * P - 3
⅓P - 3
This means that the pages read by John is ⅓P - 3
This implies that the equation to determine the number of pages in the novel is
⅓P - 3 = 114
Solving further to get the actual number of pages;
Multiply both sides by 3
3(⅓P - 3) = 114 * 3
3 * ⅓P - 3 * 3 = 114 * 3
P - 9 = 342
Add 9 to both sides
P - 9 + 9 = 342 + 9
P = 351
Hence the number of pages is 351