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yan [13]
3 years ago
6

For a given Prandtl-Meyer expansion, the upstream Mach number is 3 and the pressure ratio across the wave is P2/P1 = 0.4. Calcul

ate teh angles of the forward and rearward mach lines of the expansion fan relative to the free-stream direction
Physics
1 answer:
loris [4]3 years ago
3 0

Answer:

The angle for the forward Mach line is 19.47°

The angle for the rearward Mach line is 5.21°

Explanation:

From table A-1 (Modern Compressible Flow: with historical perspective):

\frac{P_{o} }{P_{1} } =36.73 (M₁ = 3)

If Po₁ = Po₂

\frac{P_{o2} }{P_{2} } =\frac{P_{o1} }{P_{1} } *\frac{P_{1} }{P_{2} } =36.73*\frac{1}{4} =91.825

Table A-1:

\frac{P_{o2} }{P_{2} } =91.825,M_{2} =3.63

Table A-5:

v₁ = 49.76°

μ₁ = 19.47°

v₂ = 60.55°

μ₂ = 16°

θ = 60.55 - 49.76 = 10.79°

The angle for the forward Mach line is:

μ₁ = 19.47°

The angle for the rearward Mach line is:

θr = μ₂ - θ = 16 - 10.79 = 5.21°

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problems like this A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet fi
Ymorist [56]

This question is incomplete the complete question is

A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet first into a pool. She starts with a velocity of 4.00 m/s and her takeoff point is 1.80 m above the pool. (a) What is her highest point above the board? (b) How long a time are her feet in the air? (c) What is her velocity when her feet hit the water?

Answer:

(a) Xs=0.459m

(b) t=0.984 s

(c) Vc=6.65 m/s

Explanation:

(a) To reach maximum distance

g=-9.8m/s^{2}\\ Vf=0\\v_{b}^{2}=v_{a}^{2}+2gx_{s} \\  x_{s}=\frac{0-(3^{2} )}{-2*9.8}\\ x_{s}=0.459m

(b) For Time

To find t we must find t1 and t2

as

t=t1+t2

For T1

t_{1}=(Vb-Va)/g \\t_{1}=(0-3)/9.8\\t_{1}=0.306s

For T2

x_{l}=Vbt+(1/2)gt_{2}^{2}\\   as\\x_{l}=x_{1}+x_{s}\\x_{l}=1.8+0.459\\x_{l}=2.259\\so\\t_{2}=\frac{2.259*2}{9.8} \\t_{2}=0.6789s

For Total Time

t=t1+t2

t=0.306+0.6789

t=0.984s

(c) To find Vc

Vc=Vb+gt2

Vc=(0)+(9.8)(0.6789)

Vc=6.65 m/s

7 0
2 years ago
How would the terminal velocity of a piece of tissue paper compare to the terminal velocity of a rock?
matrenka [14]

Answer: Rock require larger drag force and to achieve it rock need to move at a very high terminal velocity.  

Explanation: Terminal velocity is defined as the final velocity attained by an object falling under the gravity. At this moment weight is balanced by the air resistance or drag force and body falls with zero acceleration i.e. with a constant velocity.

Case 1: Terminal velocity of a piece of tissue paper.

The weight of tissue paper is very less and it experiences an air resistance while falling downward under the effect of gravity.

Downward gravitational force, F = mg

Upward air resistance or friction or drag force will be f_{1}

So, paper will attain terminal velocity when mg =  f_{1}

Case 2: Rock is very heavy and require larger air resistance to balance the weight of rock relative to the tissue paper case.

Downward force on rock, F = Mg

Drag force = f_{2}

Rock will attain terminal velocity when Mg = f_{2}

Mg > mg

so, f_{2} > f_{1}

And rock require larger drag force and to achieve it rock need to move at a very high terminal velocity.  

5 0
3 years ago
The town hall building is situated close to Boojho’s house. There is a clock on the top of the town hall building which rings th
KiRa [710]

Answer:

At night we experience quietness more. At this time, sounds of object will be heard more clearer

5 0
3 years ago
1) mechanical
Dafna1 [17]
I am pretty sure about these answers.Thermal goes in the 4th blank.
Mechanical goes in the 2nd blank.
Electrical goes in the 3rd blank.
I think chemical goes in the 1st blank and light goes in the 5th blank
Hope this helps
7 0
3 years ago
A projectile is launched with an initial speed of 60.0 m/s at an angle of 30.0° above the horizontal. The projectile lands on a
vodka [1.7K]

Answer:

51.96 m/s^-1

Explanation:

a) see the attachment

b) As we know the velocity of the projectile has two component, horizontal velocity v_ox. and vertical velocity v_oy as shown in the figure. At the highest point of the trajectory, the projectile has only horizontal velocity and vertical velocity is zero. Therefore at the highest point of the trajectory, the velocity of the projectile will be  

v_ox=v_o*cosФ

       =60*cos (30)

      = 51.96 m/s^-1

3 0
3 years ago
Read 2 more answers
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