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2 years ago

What is 39483048^349374*3948048/3i4u4

Engineering

1 answer:

Verizon [17]2 years ago

3 0

Answer:

1. $3.81813506×10^2^9$

2. $1.71479428×10^6^5$

3. $9.38483383×10^2^6$

4. $1.150847×10^2^9$

Explanation:

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Select the correct answer.

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The answer is A. Immediately inform her colleague

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Determine if the following errors are systematic or random. Justify your response. (a) Effect of temperature on the circuitry of

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Answer:

a) temperature: random error

b) parallax: systematic error

c) using incorrect value: systematic error

Explanation:

Systematic errors are associated with faulty calibration or reading of the equipments used and they could be avoided refining your method.

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3 years ago

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

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which then proves that ε₁+ε₂+ε₃ = 0

8 0

2 years ago

A sand has a natural water content of 5% and bulk unit weight of 18.0 kN/m3. The void ratios corresponding to the densest and lo

Zinaida [17]

Answer:

Relative density = 0.545

Degree of saturation = 24.77%

Explanation:

Data provided in the question:

Water content, w = 5%

Bulk unit weight = 18.0 kN/m³

Void ratio in the densest state, $e_{min}$ = 0.51

Void ratio in the loosest state, $e_{max}$ = 0.87

Now,

Dry density, $\gamma_d=\frac{\gamma_t}{1+w}$

$=\frac{18}{1+0.05}$

= 17.14 kN/m³

Also,

$\gamma_d=\frac{G\gamma_w}{1+e}$

here, G = Specific gravity = 2.7 for sand

$17.14=\frac{2.7\times9.81}{1+e}$

e = 0.545

Relative density = $\frac{e_{max}-e}{e_{max}-e_{min}}$

= $\frac{0.87-0.545}{0.87-0.51}$

= 0.902

Also,

Se = wG

here,

S is the degree of saturation

therefore,

S(0.545) = (0.05)()2.7

S = 0.2477

S = 0.2477 × 100% = 24.77%

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3 years ago

An engine indicator is used to determine the following: (a) speed (d) m.e.p, and IHP (e) BHP (b) temperature (c) volume of cylin

Greeley [361]

Answer:

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8 0

3 years ago