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3 years ago

Express 2/16 in thirty-seconds

Engineering

1 answer:

mafiozo [28]3 years ago

3 0

Answer:

$\frac{2}{16}$ = $\frac{4}{32}$ in thirty seconds.

Explanation:

one thirty second is one part out of 32 equal section . It is used to describe amounts accurately.

$\frac{2}{16}$ can be easily expressed as $\frac{4}{32}$

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During the recovery of a cold-worked material, which of the following statement(s) is (are) true?

Len [333]

Answer:

Some of the internal strain energy is relieved.

There is some reduction in the number of dislocations.

The electrical conductivity is recovered to its precold-worked state.

The thermal conductivity is recovered to its precold-worked state

Explanation:

The process of the recovery of a cold-worked material happens at a very low temperature, this process involves the movement and annihilation of points where there are defects, also there is the annihilation and change in position of dislocation points which leads to forming of the subgrains and the subgrains boundaries such as tilt, twist low angle boundaries.

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3 years ago

a vertical cylindrical container is being cooled in ambient air at 25 °C with no air circulation. if the initial temperature of

Sloan [31]

Answer:

the surface heat-transfer coefficient due to natural convection during the initial cooling period. = 4.93 w/m²k

Explanation:

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3 years ago

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The properties of the air in the inlet section with A1 = 0.25ab m2 in a converging-diverging channel are given as U1 = 25a,b m/s

NeX [460]

Answer:

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Explanation:

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3 years ago

A gas tank is known to have a thickness of 0.5 inches and an internal pressure of 2.2 ksi. Assuming that the maximum allowable s

sergiy2304 [10]

Answer:

$D_o=11.9inch$

Explanation:

From the question we are told that:

Thickness $T=0.5$

Internal Pressure $P=2.2Ksi$

Shear stress $\sigma=12ksi$

Elastic modulus $\gamma= 35000$

Generally the equation for shear stress is mathematically given by

$\sigma=\frac{P*r_1}{2*t}$

Where

r_i=internal Radius

Therefore

$12=\frac{2.2*r_1}{2*0.5}$

$r_i=5.45$

Generally

$r_o=r_1+t$

$r_o=5.45+0.5$

$r_o=5.95$

Generally the equation for outer diameter is mathematically given by

$D_o=2r_o$

$D_o=11.9inch$

Therefore

Assuming that the thin cylinder is subjected to integral Pressure

Outer Diameter is

$D_o=11.9inch$

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3 years ago

Indicate the correct statement about the effect of Reynolds number on the character of the flow over an object.

sergeinik [125]

Answer:

If Reynolds number increases the extent of the region around the object that is affected by viscosity decreases.

Explanation:

Reynolds number is an important dimensionless parameter in fluid mechanics.

It is calculated as;

$R_e__N} = \frac{\rho vd}{\mu}$

where;

ρ is density

v is velocity

d is diameter

μ is viscosity

All these parameters are important in calculating Reynolds number and understanding of fluid flow over an object.

In aerodynamics, the higher the Reynolds number, the lesser the viscosity plays a role in the flow around the airfoil. As Reynolds number increases, the boundary layer gets thinner, which results in a lower drag. Or simply put, if Reynolds number increases the extent of the region around the object that is affected by viscosity decreases.

5 0

3 years ago