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SVETLANKA909090 [29]
3 years ago
10

Some_____

Engineering
1 answer:
Thepotemich [5.8K]3 years ago
5 0

Answer:

it’s IGS

Explanation:

because i read

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When the outside temperature is 5.2 ⁰C, a steel beam of cross-sectional area 52 cm2 is installed in a building with the ends of
il63 [147K]

Multiply the coefficient by the change in temperature:

1.1*10^-5 x (37-5.2) = 0.0003498

Multiply Young's modulus by the area by the above answer:

2*10^11 x 52 * 0.0003498 x (1/100)^2 = 3.63792 x 10^5 N

6 0
3 years ago
The air velocity in the duct of a heating system is to be measured by a Pitot-static probe inserted into the duct parallel to th
Ilia_Sergeevich [38]

Answer:

Flow velocity

50.48m/s

Pressure change at probe tip

1236.06Pa

Explanation:

Question is incomplete

The air velocity in the duct of a heating system is to be measured by a Pitot-static probe inserted into the duct parallel to the flow. If the differential height between the water columns connected to the two outlets of the probe is 0.126m, determine (a) the flow velocity and (b) the pressure rise at the tip of the probe. The air temperature and pressure in the duct are 352k and 98 kPa, respectively

solution

In this question, we are asked to calculate the flow velocity and the pressure rise at the tip of probe

please check attachment for complete solution and step by step explanation

8 0
3 years ago
Here you go!!!!!!!!!!!!!!!!!1
sweet [91]

Answer:

Im confused, what does this mean

Explanation:

i mean, thx lol

3 0
3 years ago
or a metal pipe used to pump tomato paste, the overall heat- transfer coefficient based on internal area is 2 W/(m2 K). The insi
igomit [66]

Answer: ok the best one would be letter s because it goes

Explanation:

467,,mm tubing should do

7 0
3 years ago
The fan blades suddenly experience an angular acceleration of 2 rad/s2. If the blades are rotating with an initial angular veloc
madreJ [45]

Answer:

Option B

116 ft/s^{2}

Explanation:

\theta=2 rev=2(2\pi)=4\pi

\alpha \theta=0.5(\omega_f^{2}-\omega_i^{2})

\alpha (4\pi)= 0.5(\omega_f^{2}-\omega_i^{2})

\alpha (8\pi)= (\omega_f^{2}-\omega_i^{2})

(2) (8\pi)= (\omega_f^{2}-\omega_i^{2})

(2) (8\pi)= (\omega_f^{2}-4^{2})

\omega_f=8.14 rads/s

v=r\omega=1.75*8.14=14.245 ft/s

Centripetal acceleration =\omega_f^{2} r=8.14^{2}*1.75=115.95 ft/s^{2}

Tangential component=dr=2*1.75=3.5

Resultant=\sqrt{3.5^{2}+115.95^{2}}\approx 116 ft/s^{2}

5 0
3 years ago
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