Lick the bullet and push it down
Answer:
σ = 391.2 MPa
Explanation:
The relation between true stress and true strain is given as:
σ = k εⁿ
where,
σ = true stress = 365 MPa
k = constant
ε = true strain = Change in Length/Original Length
ε = (61.8 - 54.8)/54.8 = 0.128
n = strain hardening exponent = 0.2
Therefore,
365 MPa = K (0.128)^0.2
K = 365 MPa/(0.128)^0.2
k = 550.62 MPa
Now, we have the following data:
σ = true stress = ?
k = constant = 550.62 MPa
ε = true strain = Change in Length/Original Length
ε = (64.7 - 54.8)/54.8 = 0.181
n = strain hardening exponent = 0.2
Therefore,
σ = (550.62 MPa)(0.181)^0.2
<u>σ = 391.2 MPa</u>
Answer:
2)
3) 
Explanation:
1) Expressing the Division as the summation of the quotient and the remainder
for
118, knowing it is originally a decimal form:
118:2=59 +(0), 59/2 =29 + 1, 29/2=14+1, 14/2=7+0, 7/2=3+1, 3/2=1+1, 1/2=0+1

2) 
Similarly, we'll start the process with the absolute value of -49 since we want the positive value of it. Then let's start the successive divisions till zero.
|-49|=49
49:2=24+1, 24:2=12+0,12:2=6+0,6:2=3+0,3:2=1+1,1:2=0+1
100011

3) 
The first step on that is dividing by 16, and then dividing their quotient again by 16, so on and adding their remainders. Simply put:

A short framing member that fills the space between the rough sill and the soleplate is a cripple stud.
(HAVE A GOOD DAY!!!)
Answer:
a)
, b) 
Explanation:
The Brinell hardness can be determined by using this expression:

Where
and
are the indenter diameter and the indentation diameter, respectively.
a) The Brinell hardness is:
![HB = \frac{2\cdot (1000\,kgf)}{\pi\cdot (10\,mm)^{2}} \cdot \left[\frac{1}{1-\sqrt{1-\frac{(2.50\,mm)^{2}}{(10\,mm)^{2}} } } \right]](https://tex.z-dn.net/?f=HB%20%3D%20%5Cfrac%7B2%5Ccdot%20%281000%5C%2Ckgf%29%7D%7B%5Cpi%5Ccdot%20%2810%5C%2Cmm%29%5E%7B2%7D%7D%20%5Ccdot%20%5Cleft%5B%5Cfrac%7B1%7D%7B1-%5Csqrt%7B1-%5Cfrac%7B%282.50%5C%2Cmm%29%5E%7B2%7D%7D%7B%2810%5C%2Cmm%29%5E%7B2%7D%7D%20%7D%20%7D%20%20%5Cright%5D)
b) The diameter of the indentation is obtained by clearing the corresponding variable in the Brinell formula:

![d = (10\,mm)\cdot\sqrt{1-\left[1-\frac{2\cdot (500\,kgf)}{\pi\cdot (300)\cdot (10\,mm)^{2}} \right]^{2}}](https://tex.z-dn.net/?f=d%20%3D%20%2810%5C%2Cmm%29%5Ccdot%5Csqrt%7B1-%5Cleft%5B1-%5Cfrac%7B2%5Ccdot%20%28500%5C%2Ckgf%29%7D%7B%5Cpi%5Ccdot%20%28300%29%5Ccdot%20%2810%5C%2Cmm%29%5E%7B2%7D%7D%20%20%5Cright%5D%5E%7B2%7D%7D)
