Answer:
The answer is "22.501,-22.899"
Explanation:
Just as in the previous problems find the angle the velocity makes with the x-axis and radius of curvature.
[tex]x= 2t^2 + 3t — 1\\\\y=5t-2\\\\x=4t+3\\\\y=5\\\\\tan \alpha (t = 1) =\frac{y}{x}=\frac{5}{4+3}=\frac{5}{7} \to alpha=35.54^{\circ}\\\\[/tex]
For the radius of curvature, we can use the expression from the last two problems, but first express the position and derivatives as y(x).
[tex]y(x)=2(\frac{y+2}{5})^2+3(\frac{y+2}{5})-1=\frac{1}{25}(2y^2+23y+13)\\\\y'(x)=\frac{1}{25}(4y+23)\\\\y''(x)=\frac{4}{25}\\\\\rho(t=1)=\frac{[1+(\frac{dy}{dx})^2]^{\frac{3}{2}}}{\frac{d^2y}{dx^2}}=\frac{(1+(\frac{35}{25})^2)^{\frac{3}{2}}}{4}25=31.828[/tex]
The position for the center of the radius of curvature [tex]\vec{r}[/tex], (finding this expression is easy and is left as an exercise for the reader.)
[tex]\to \vec{r} = \hat{x}(x + \rho \sin \alpha) + \hat{y}(y- \rho \cos \alpha)\\\\= (4 + 18.501, 3-25.899)\\\\=(22.501, -22.899)[/tex]
Calculate the biaxial stresses σ1 and σ2 for the biaxial stress case, where ε1 = .0020 and ε2 = –.0010 are determined experimentally on an aluminum member of elastic constants, E = 71 GPa and v = 0.35. Also, determine the value for the maximum shear stress.
Answer:
i) σ1 = 133.5 MPa
σ2 = -2427 MPa
ii) 78.89 MPa
Explanation:
Given data:
ε1 = 0.0020 and ε2 = –0.0010
E = 71 GPa
v = 0.35
i) Determine the biaxial stresses σ1 and σ2 using the relations below
ε1 = σ1 / E - v (σ2 / E) -----( 1 )
ε2 = σ2 / E - v (σ1 / E) -------( 2 )
resolving equations 1 and 2
σ1 = E / 1 - v^2 { ε1 + vε2 } ---- ( 3 )
σ2 = E / 1 - v^2 { ε2 + vε1 } ----- ( 4 )
input the given data into equation 3 and equation 4
σ1 = 133.5 MPa
σ2 = -2427 MPa
ii) Calculate the value of the maximum shear stress ( Zmax )
Zmax = ( σ1 - σ2 ) / 2
= 133.5 - ( - 2427 ) / 2
= 78.89 MPa
The number of pulses per second from IGBTs is referred to as
: Một nền kinh tế có cấu trúc như sau:
C = 80 + 0,8(Y - T); T = 100 ;
I = 130; G = 120;
MSr = MS/CPI = 200;
MD = 0,2Y – 10i
Yêu cầu:
1. Xác định thu nhập và lãi suất cân bằng?
2. Muốn sản lượng cân bằng tăng 500 thì chính phủ cần thay đổi thuế như thế nào?
3. Liệu mục tiêu ở câu 2 có thể đạt đựơc bằng chính sách tiền tệ hay không? Tại sao?
Answer:
Haha I'm a great guy but my friend has been in a day of the day and a lot to be able and I'm happy holi and a lot to the world of the day and day to
Determine the voltage which must be applied to a 1 k 2 resistor in order that a current of
10 mA may flow.
Answer:
The correct solution is "20 volt".
Explanation:
Given that:
Current,
I = 10 mA
or,
= [tex]10\times 10^{-3} \ A[/tex]
Resistance,
R = 2 K ohm
or,
= [tex]1\times 10^3 \ ohm[/tex]
Now,
The voltage will be:
⇒ [tex]V=IR[/tex]
By putting the values, we get
[tex]=10\times 10^{-3}\times 2\times 10^{3}[/tex]
[tex]= 20 \ volt[/tex]
Type the correct answer in each box. Spell all words correctly. According to the priority matrix, which tasks should an entrepreneur complete first? According to the priority matrix, entrepreneurs should first complete tasks that are blank and important.
Answer:
Development of creative and develop ideas
Explanation:
First task as an entrepreneur is to be creative and develop ideas. The person must design the product based on which he will develop the business strategy.
The remaining activities such as marketing, fund raising, recruitment etc. comes at a later stage.
A brittle material is subjected to a tensile stress of 1.65 MPa. If the specific surface energy and modulus of elasticity for this material are 0.60 J/m2 and 2.0 GPa, respectively. What is the maximum length of a surface flaw that is possible without fracture
Answer:
The maximum length of a surface flaw that is possible without fracture is
[tex]2.806 \times 10^{-4} m[/tex]
Explanation:
The given values are,
σ=1.65 MPa
γs=0.60 J/m2
E= 2.0 GPa
The maximum possible length is calculated as:
[tex]\begin{gathered}a=\frac{2 E \gamma_{s}}{\pi \sigma^{2}}=\frac{(2)\left(2 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\right)(0.60 \mathrm{~N} / \mathrm{m})}{\pi\left(1.65\times 10^{6} \mathrm{~N} / \mathrm{m}^{2}\right)^{2}} \\=2.806 \times 10^{-4} \mathrm{~m}\end{gathered}[/tex]
The maximum length of a surface flaw that is possible without fracture is
[tex]2.806 \times 10^{-4} m[/tex]
