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mars1129 [50]
3 years ago
15

What are the partial products of 2.3 x 2.6

Engineering
1 answer:
kicyunya [14]3 years ago
7 0
Answer: 5.98
2.3 x 2.6 = 5.98
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A circular bar is 800mm in length and 32mm in diameter. The bar is made from a material with a modulus of elasticity E = 150 GPa
Yuki888 [10]

Answer:

For any material if ∈ is the axial strain then the lateral strain is given by -μ∈ is the lateral strain in the object

Where,

μ is the poisson's ratio of the material

The longitudinal strain is calculated as follows

\varepsilon _{axial}=\frac{\Delta length}{Length_{original}}\\\\\therefore \varepsilon _{axial}=\frac{0.7}{800}=8.75\times 10^{-4}

Thus the lateral strain becomes

\varepsilon _{lateral}=-\mu\varepsilon _{axial}\\\\\varepsilon _{lateral}=-0.27\times 8.75\times 10^{-4}=-2.36\times 10^{-4}

now by definition of lateral strain we have

\varepsilon _{lateral}=\frac{\Delta diameter}{diameter_{original}}\\\\\Rightarrow \Delta Diameter=-2.36\times 10^{-4}\times 32=-7.56\times 10^{-3}\\\\D_{f}-D_{i}=-7.56\times 10^{-3}\\\\D_{f}=32-7.56\times 10^{-3}=31.992mm

By hookes law the stress developed due to the given strain is given by

\sigma =\varepsilon _{axial}E

Applying values we get

\sigma =8.75\times 10^{-4}\times 150\times 10^{9}\\\\\sigma =131.25MPa

Thus the force is calculated as

Force=\sigma \times Area\\\\Force=131.25\times 10^{6}\times \frac{\pi (32\times 10^{-3})^{4}}{4}\\\\Force=105.55kN

5 0
3 years ago
Which of the following is not an example of heat generation? a)- Exothermic chemical reaction in a solid b)- Endothermic Chemica
Sav [38]

Answer:

b) Endothermic Chemical Reactions in a solid

Explanation:

Endothermic reactions consume energy, which will result in a cooler solid when the reaction finishes.

8 0
3 years ago
The equation for the velocity V in a pipe with diameter d and length L, under laminar condition is given by the equation V=Δpdsq
Citrus2011 [14]

Answer:

Given that

V=\dfrac{\Delta Pd^2}{32\mu L}

LHS of above given equation have dimension [M^oL^{1}T^{-1}].

Now find the dimension of RHS

Dimension of P = [ML^{-1}T^{-2}].

Dimension of d=  [M^{0}L^{1}T^{0}].

Dimension of μ =  [ML^{-1}T^{-1}].

Dimension of L=  [M^{0}L^{1}T^{0}].

So

\dfrac{\Delta Pd^2}{32\mu L}=\dfrac{[ML^{-1}T^{-2}].[M^{0}L^{1}T^{0}]^2}{[ML^{-1}T^{-1}].[M^{0}L^{1}T^{0}]}

\dfrac{\Delta Pd^2}{32\mu L}=[M^0L^{1}T^{-1}]

It means that both sides have same dimensions.

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3 years ago
Just getting to your class is enough for showing up.<br> A) True<br> B) False
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False you have to do the work and perricapate
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4 years ago
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