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Aleks04 [339]
2 years ago
10

Prepare deflection angles for staking the curves in problem 2-2. In every example assume that Sta. P.I. = 37+53.21. Carry the re

sults to the precision used in section 2-9. (a) Set full stations. Partial answers: Sta. T.C. = 32+58.83; C.T. = 42+40.83; Deflection T.C. to 33+00 = 0°20.58'.
Engineering
1 answer:
weeeeeb [17]2 years ago
4 0

Answer:

ya mom

Explanation:

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The future goal of health information is appropriate access to ubiquitous, personalized, specific information relevant to the si
erastovalidia [21]
I thinks it’s a A.True
3 0
3 years ago
Define the overall heat transfer coefficient.
9966 [12]

Answer and Explanation:

In thermodynamics, the overall heat transfer coefficient also referred as film effectiveness is a constant of proportionality between force drive for the heat flow and the heat flux.

It gives the measure of the heat transfer as a result of  convection or conduction. The coefficient of overall heat transfer depends on surface area, resistance of the material, temperature difference, thickness, etc.

It is given by:

Q = UA\Delta T

where

U = overall heat transfer coefficient

Its SI units is W/m^{2}K.

6 0
3 years ago
.) If the charges attracting each other in the problem above have equal magnitude, what is the magnitude of each charge?
Sedaia [141]

Answer:

Not seeing any other information, the best answer I can give is 2m.

Explanation:

M = magnitude

You see, if they have an equal charge, and you add them, it'd be 2 * m, or 2m.

8 0
3 years ago
water flows in a horizontal constant-area pipe; the pipe diameter is 75 mm and the average flow speed is 5 m/s. At the pipe inle
Veronika [31]

Answer:

Head loss = 28.03 m

Explanation:

According to Bernoulli's theorem for fluids  we have

\frac{P}{\gamma _{w}}+\frac{V^{2}}{2g}+z=Constant

Applying this between the 2 given points we have

\frac{P_{1}}{\gamma _{w}}+\frac{V_{1}^{2}}{2g}+z_{1}=\frac{P_{2}}{\gamma _{w}}+\frac{V_{2}^{2}}{2g}+z_{2}+h_{l}

Here h_{l} is the head loss that occurs

\therefore h_{l}=\frac{P_{1}}{\gamma _{w}}+\frac{V_{1}^{2}}{2g}+z_{1}-\frac{P_{2}}{\gamma _{w}}-\frac{V_{2}^{2}}{2g}-z_{2}

Since the pipe is horizantal we have z_{1}-z_{2}=0

Applying contunity equation between the 2 sections we get

A_{1}V_{1}=A_{2}V_{2}\\\\\therefore V_{1}=V_{2}(\because A_{1}=A_{2})

Since the cross sectional area of the both the sections is same thus the speed

is also same

Using this information in the above equation of head loss we obtain

h_{l}=\frac{1}{\gamma _{w}}(P_{1}-P_{2})

Applying values we get

h_{l}=\frac{1}{9810}\times (275\times 10^{3})m\\\\\therefore h_{l}=28.03m

3 0
3 years ago
When is the output of an XOR gate HIGH? explain​
erma4kov [3.2K]

Answer:

The output of a NOR gate is LOW whenever one or more inputs are HIGH. The output of an XOR gate is HIGH whenever the two inputs are different. The output of an XNOR gate is HIGH whenever the two inputs are identical

5 0
2 years ago
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