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Vesna [10]
3 years ago
14

Drawings or blank) can also provide additional information that may be difficult to convey in writing, and they can help readers

picture in their heads what they are reading.
please please please help
Engineering
2 answers:
Andrews [41]3 years ago
8 0

Answer: A is the answer, The use of headings in boldface type leads the reader to specific information.

Explanation:

klio [65]3 years ago
5 0
I think it’s A your welcome
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A Pitot-static probe is used to measure the speed of an aircraft flying at 3000 m. If the differential pressure reading is 3200
coldgirl [10]

Answer:

Speed of aircraft ; (V_1) = 83.9 m/s

Explanation:

The height at which aircraft is flying = 3000 m

The differential pressure = 3200 N/m²

From the table i attached, the density of air at 3000 m altitude is; ρ = 0.909 kg/m3

Now, we will solve this question under the assumption that the air flow is steady, incompressible and irrotational with negligible frictional and wind effects.

Thus, let's apply the Bernoulli equation :

P1/ρg + (V_1)²/2g + z1 = P2/ρg + (V_2)²/2g + z2

Now, neglecting head difference due to high altitude i.e ( z1=z2 ) and V2 =0 at stagnation point.

We'll obtain ;

P1/ρg + (V_1)²/2g = P2/ρg

Let's make V_1 the subject;

(V_1)² = 2(P1 - P2)/ρ

(V_1) = √(2(P1 - P2)/ρ)

P1 - P2 is the differential pressure and has a value of 3200 N/m² from the question

Thus,

(V_1) = √(2 x 3200)/0.909)

(V_1) = 83.9 m/s

4 0
3 years ago
2. A trapezoidal channel has a bottom width of 4 m and side slopes of 2:1 (H:V). If the flow rate is 50 m3/s at a depth of 3.6 m
Daniel [21]

Answer:

Explanation:

Sum of the side slope = 2 + 1 = 3

Length of first slope = 2/3 X 3.6 = 2 X 1.2 = 2.4m

Lenght of second slope = 1/3 X 3.6 = 1.2m

Area of the trapezoidal channel = (2.4 + 1.2)/2 X 3.6 = 1.8 X 3.6 = 6.48m²

Alternate dept = 50m³/6.48m²= 7.716m

4 0
3 years ago
True or False; The Neutrons in an atom have a neutral charge.​
yanalaym [24]

Answer:

true

Explanation:

if it is not true it is false

3 0
3 years ago
Read 2 more answers
A 3-phase induction motor with 4 poles is connected to a voltage source with an amplitude of 209 Vrms and a frequency of 120 Hz.
Ket [755]

Answer:

<em>T = 25.41 Nm</em>

Explanation:

Calculating Nsync (Synchronous Speed):

Nsync = 120f/P

Nsync = 120 x 120 / 4\\Nsync = 3600 rpm

Wsync = 3600 * 2\pi /180\\Wsync = 377 rad/s

Calculating s (Slip):

s = (Nsync - Nm) / Nsync \\[tex]s = (3600-2464)/3600\\s = 0.3156

Calculating Vth (Thevenin Voltage):

Vth = Vph (Xm / \sqrt{Rs^{2} + (Xs+Xm)^{2}  })\\Vth = 209 (76 / \sqrt{(2.3)^{2} + (1.3 + 76)^{2}  }\\Vth = 205.39 V

Calculating Rth (Thevenin Resistance):

Rth = Rs (Xm/Xs + Xm)^{2} \\Rth = 2.3 (76/1.3 + 76)^{2} \\Rth = 2.22 ohm

Calculating Xth (Thevenin Reactance):

Xth = Xs = 1.3 ohm

Calculating Torque:

T = (3Vth^{2}Rr/s) / (Wsync[(Rth+Rr/s)^{2} + (Xth + Xr)^{2}])\\T = (3*205.39*0.7/0.3156) / 377[(2.22+0.7/0.315)^{2} + (1.3+1.8)^{2}]

T = 280699 / 377 [19.69 + 9.61]

<em>T = 25.41 Nm</em>

4 0
3 years ago
2
aksik [14]

Answer:tech A

Explanation:

5 0
3 years ago
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