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anzhelika [568]
2 years ago
8

Why dose bob not let humans touch him one and only Ivan

Engineering
1 answer:
gogolik [260]2 years ago
8 0
Sorry man i don’t know sorry
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Why do computers use 1 and 0
tangare [24]

I like ya cut g

Explanation:

bonK no Uhm

6 0
3 years ago
Read 2 more answers
Consider a cubic workpiece of rigid perfect plastic material with side length lo. The cube is deformed plastically to the shape
Taya2010 [7]

Answer:  ε₁+ε₂+ε₃ = 0

Explanation: Considering the initial and final volume to be constant which gives rise to the relation:-

                         l₀l₀l₀=l₁l₂l₃

                        \frac{lo*lo*lo}{l1*l2*l3}=1.0

                      taking natural log on both sides

                              ln(\frac{(lo*lo*lo)}{l1*l2*l3})=ln(1)

Considering the logarithmic Laws of division and multiplication :

                                ln(AB) = ln(A)+ln(B)

                                ln(A/B) = ln(A)-ln(B)

                           ln(\frac{(l1)}{lo})*ln(\frac{(l2)}{lo})*ln(\frac{(l3)}{lo}) = 0

Use the image attached to see the definition of true strain defined as

                         ln(l1/1o)= ε₁

which then proves that ε₁+ε₂+ε₃ = 0

8 0
3 years ago
a ________________ is the flat, low-lying area adjacent to a stream channel that receives overbank flow
Umnica [9.8K]

Answer:

flood plain

Explanation:

it should be called a flood plain which makes sense because it’s not a deposition.

5 0
2 years ago
A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 30.00 a
kirill [66]

Answer:

105.70 mm

Explanation:

Poisson’s ratio, v is the ratio of lateral strain to axial strain.

E=2G(1+v) where E is Young’s modulus, v is poisson’s ratio and G is shear modulus

Since G is given as 25.4GPa, E is 65.5GPa, we substitute into our equation to obtain poisson’s ratio

\begin{array}{l}\\65.5{\rm{ GPa}} = 2\left( {25.4{\rm{ GPa}}} \right)(1 + \upsilon )\\\\\upsilon = 0.2893\\\end{array}

Original length L_(i}

\upsilon = - \left( {\frac{{\left( {\frac{{{d_f} - {d_i}}}{{{d_i}}}} \right)}}{{\left( {\frac{{{L_f} - {L_i}}}{{{L_i}}}} \right)}}} \right)

Where d_{f} is final diameter, d_{i} is original diameter, L_{f} is final length and L_{i} is original length.

\begin{array}{l}\\0.2893 = - \left( {\frac{{\left( {\frac{{30.04{\rm{ mm}} - {\rm{30 mm}}}}{{{\rm{30 mm}}}}} \right)}}{{\left( {\frac{{{\rm{105.2 mm}} - {L_i}}}{{{L_i}}}} \right)}}} \right)\\\\\left( {\frac{{{\rm{105.2 mm}} - {L_i}}}{{{L_i}}}} \right) = - 4.6088 \times {10^{ - 3}}\\\end{array}

\begin{array}{l}\\105.2 - {L_i} = - \left( {4.6088 \times {{10}^{ - 3}}} \right){L_i}\\\\105.2 = 0.9953{L_i}\\\\{L_i} = 105.70{\rm{ mm}}\\\end{array}

Therefore, the original length is 105.70 mm

7 0
4 years ago
A direct-coupled amplifier has a low-frequency gain of 40 dB, poles at 2 MHz and 20 MHz, a zero on the negative real axis at 200
rewona [7]

Answer:

Explanation:

Low frequency gain is= 40db= 20logK=>100 poles at 2MHz,20MHz

Zero at -200MHz, zero at infinity.

A) A(s) = 100FH(s)

B) Poles (1): 2 pi × 2 × 10^6= 4pi × 10^6MHz

              (2): 2pi × 20 × 10^6= 4pi × 10^6 MHz

 Zeroed: 2pi × 10^6 × 200= 400pi × 10^6, at infinity.

T/(S) = (1 + S/400π × 10^6)/S(1 + S/4π × 10^6)(1 + S/4π × 10^6)

8 0
3 years ago
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