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dexar [7]
3 years ago
8

Privacy settings allow account owners to decide who can

Engineering
1 answer:
VLD [36.1K]3 years ago
6 0
Access their accounts.

anyone can search for them online.
they only meet in person with who they choose.
no one can view their personal information (other than companies they give it to.)
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What is the built-in pollution control system in an incinerator called
Kobotan [32]

Explanation:

hbyndbnn☝️

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3 years ago
The best way to check the efficiency of individual cylinders is:________.
KATRIN_1 [288]

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To run. The machine one at a time

Explanation:

8 0
4 years ago
Air flows at 45m/s through a right angle pipe bend with a constant diameter of 2cm. What is the overall force required to keep t
HACTEHA [7]

Answer:

b)1.08 N

Explanation:

Given that

velocity of air V= 45 m/s

Diameter of pipe = 2 cm

Force exerted by fluid  F

F=\rho AV^2

So force exerted in x-direction

F_x=\rho AV^2

F_x=1.2\times \dfrac{\pi}{4}\times 0.02^2\times 45^2

F=0.763 N

So force exerted in y-direction

F_y=\rho AV^2

F_y=1.2\times \dfrac{\pi}{4}\times 0.02^2\times 45^2

F=0.763 N

So the resultant force R

R=\sqrt{F_x^2+F_y^2}

R=\sqrt{0.763^2+0.763^2}

R=1.079

So the force required to hold the pipe is 1.08 N.

3 0
3 years ago
A stainless steel ball (rho = 8055 kg/m3, cp = 480 J/kg·K) of diameter D = 0.21 m is removed from the oven at a uniform temperat
Nataliya [291]

Answer:

Explanation:

The complete detailed  explanation which answer the question efficiently is shown in the attached files below.

I hope it helps a lot !

5 0
3 years ago
Read 2 more answers
A 1-ft rod with a diameter of 0.5 in. is subjected to a tensile force of 1,300 lb and has an elongation of 0.009 in. The modulus
iragen [17]

Answer:

E = 8.83 kips

Explanation:

First, we determine the stress on the rod:

\sigma = \frac{F}{A}\\\\

where,

σ = stress = ?

F = Force Applied = 1300 lb

A = Cross-sectional Area of rod = 0.5\pi \frac{d^2}{4} = \pi \frac{(0.5\ in)^2}{4} = 0.1963\ in^2

Therefore,

\sigma = \frac{1300\ lb}{0.1963\ in^2} \\\\\sigma = 6.62\ kips

Now, we determine the strain:

strain = \epsilon = \frac{elongation}{original\ length} \\\\\epsilon = \frac{0.009\ in}{12\ in}\\\\\epsilon =  7.5\ x\ 10^{-4}

Now, the modulus of elasticity (E) is given as:

E = \frac{\sigma}{\epsilon}\\\\E = \frac{6.62\ kips}{7.5\ x\ 10^{-4}}

<u>E = 8.83 kips</u>

7 0
3 years ago
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