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goblinko [34]
3 years ago
8

Consider a rectangular wing mounted in a low-speed subsonic wind tunnel. The wing model completely spans the test-section, so th

at the flow "sees" essentially an infinite wing. The wing has a NACA 23012 airfoil section and a chord of 0.23 m, where the lift on the entire wing is measured as 200 N by the wind tunnel force balance. Determine the angle of attack if the wing span, airflow pressure, temperature, and velocity are 2 m, 1 atm, 303 K, and 42 m/s, respectively. Refer to the Appendix graphs given below for standard values. (Round the final answer to the nearest whole number.)
Physics
1 answer:
Ksivusya [100]3 years ago
3 0

Answer:

Answer for the question :

"Consider a rectangular wing mounted in a low-speed subsonic wind tunnel. The wing model completely spans the test-section, so that the flow "sees" essentially an infinite wing. The wing has a NACA 23012 airfoil section and a chord of 0.23 m, where the lift on the entire wing is measured as 200 N by the wind tunnel force balance. Determine the angle of attack if the wing span, airflow pressure, temperature, and velocity are 2 m, 1 atm, 303 K, and 42 m/s, respectively. Refer to the Appendix graphs given below for standard values" is in the attachment.

Explanation:

Download pdf
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A piston–cylinder assembly fitted with a slowly rotating paddle wheel contains 0.13 kg of air, initially at 300 K. The air under
olganol [36]

Answer:

Explanation:

From the first law of thermodynamics

Q=ΔU+W

Q=heat supplies to the system

ΔU=change in internal energy of the system

W= work done by the gas  by the system

Q =12 KJ

ΔU=U_{2} -U_{1}=(mR/k-1)ΔT

ΔU=\frac{0.3*0.287}{1.4-1}*(460-400)

     =14.924 KJ

Q=ΔU+W

W=-2.924 KJ

W_{net}=W_{air}+W_{paddle}

W_{air}=work done by the air

W_{air}= P(V2-V1)

W_{air}=mRT2-mRT1

W_{air}=mR(T2-T1)=0.138*0.287*(460-300)=

                           =6.33 KJ

W_{net}=W_{air}+W_{paddle}

                      =-2.924-6.33

W_{paddle}=-9.26  KJ                    

7 0
3 years ago
A technician wearing a brass bracelet enclosing area 0.00500 m2 places her hand in a solenoid whose magnetic field is 3.10 T dir
aliya0001 [1]

Explanation:

Given that,

Area enclosed by a brass bracelet, A=0.005\ m^2

Initial magnetic field, B_i=3.1\ T

The electrical resistance around the circumference of the bracelet is, R = 0.02 ohms

Final magnetic field, B_f=0.93\ T

Time, t=16\ ms=16\times 10^{-3}\ s

The expression for the induced emf is given by :

\epsilon=-\dfrac{d\phi}{dt}

\phi = magnetic flux

\epsilon=-\dfrac{d(BA)}{dt}

\epsilon=-A\dfrac{d(B)}{dt}

\epsilon=-A\dfrac{B_f-B_i}{t}

\epsilon=-0.005 \times \dfrac{0.93-3.1}{16\times 10^{-3}}

\epsilon=0.678\ volts

So, the induced emf in the bracelet is 0.678 volts.

Using ohm's law to find the induced current as :

V = IR

I=\dfrac{V}{R}

I=\dfrac{0.678}{0.02}

I = 33.9 A

or

I = 34 A

So, the induced current in the bracelet is 34 A. Hence, this is the required solution.

5 0
3 years ago
A student is building a simple circuit with a battery, light bulb, and copper wires. When she connects the wires to the battery
denis23 [38]
It’s will be B because the circuit had a open or close so if that doesn’t work than it’s because it’s open
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4 years ago
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Rate in miles/hours if your run at a speed 2.2 20miles
Vinvika [58]
9.1 miles per hour because 2.2 is your hours right?
5 0
4 years ago
Calculate the acceleration of a bus full speed changes from 6 miles to 12 miles over a period of three seconds
Jet001 [13]

Answer:

0.90m/s²

Explanation:

Given parameters:

Initial speed  = 6miles/hr

Final speed  = 12miles/hr

Time taken = 3 seconds

Unknown:

Acceleration = ?

Solution:

Acceleration is the rate of change of velocity with time. It is mathematically given as:

         Acceleration  = \frac{Final speed - Initial speed}{Time}

We need to convert miles/hr to meters/seconds

Initial speed = 6 x \frac{miles }{hour} x \frac{1hr}{3600s}  x \frac{1609.34m}{1mile}

                      = 2.68m/s

Final speed = 12 x \frac{1609.34}{3600}  = 5.37m/s

      Acceleration  = \frac{5.37  - 2.68}{3}   = 0.90m/s²

4 0
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