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3 years ago

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In the flow past a compression corner, the upstream Mach number and pressure are 3.5 and 1 atm, respectively. Downstream of the

corner, the pressure is 5.48 atm. Calculate the deflection angle of the corner.

1 answer:

yulyashka [42]3 years ago

4 0

Answer:

$\theta=23.7^{\circ}$

Explanation:

The ratio of pressure 2 to 1 us 5.48/1= 5.48 rounded off as 5.5.

Referring to table A.2 of modern compressible flow then $M_{\beta_1}=2.2$

Also

$M_{\beta_1}=M_1 sin \beta$ and making $sin\beta$ the subject of the formula then

$sin\beta=\frac {2.2}{3.5}\\\beta=38.94^{\circ}$

Making reference to $\theta-\beta-M$ diagram then

$\theta=23.7^{\circ}$

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Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consid

Korolek [52]

Answer:

a

$z= 2.5 \ m$

b

$z = (1 \ m , 4 \ m )$

Explanation:

From the question we are told that

Their distance apart is $d = 5.00 \ m$

The wavelength of each source wave $\lambda = 6.0 \ m$

Let the distance from source A where the construct interference occurred be z

Generally the path difference for constructive interference is

$z - (d-z) = m \lambda$

Now given that we are considering just the straight line (i.e points along the line connecting the two sources ) then the order of the maxima m = 0

so

$z - (5-z) = 0$

=> $2 z - 5 = 0$

=> $z= 2.5 \ m$

Generally the path difference for destructive interference is

$|z-(d-z)| = (2m + 1)\frac{\lambda}{2}$

=> $|2z - d |= (0 + 1)\frac{\lambda}{2}$

=> $|2z - d| =\frac{\lambda}{2}$

substituting values

$|2z - 5| =\frac{6}{2}$

=> $z = \frac{5 \pm 3}{2}$

So

$z = \frac{5 + 3}{2}$

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and

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3 years ago

If v = 5.00 meters/second and makes an angle of 60° with the negative direction of the y–axis, calculate all possible values of

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Vx = + 4.33 m/s. Hope this helps

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2 years ago

Heather and Jerry are standing on a bridge 49 mm above a river. Heather throws a rock straight down with a speed of 17 m/sm/s .

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Answer:

3.467 s

Explanation:

given,

distance , d = 49 mm = 0.049 m

initial speed of the of the rock, v = 17 m/s

time taken by the Heather rock to reach water

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$s = ut +\dfrac{1}{2}at^2$

taking downward as negative

$-0.049 = -17 t -\dfrac{1}{2}\times 9.8\times t^2$

4.9 t² + 17 t - 0.049 = 0

now,

$t_1 = \dfrac{-(17)\pm \sqrt{17^2 - 4\times 4.9 \times (-0.049)}}{2\times 4.9}$

t₁ = -3.47 s , 0.0028 s

rejecting negative values

t₁ = 0.0028 s

now, time taken by the ball of Jerry

using equation of motion

$s = ut +\dfrac{1}{2}at^2$

taking downward as negative

$-0.049 = 17 t -\dfrac{1}{2}\times 9.8\times t^2$

4.9 t² - 17 t - 0.049 = 0

now,

$t_2 = \dfrac{-(-17)\pm \sqrt{17^2 - 4\times 4.9 \times (-0.049)}}{2\times 4.9}$

t₂ = 3.47 s ,-0.0028 s

rejecting negative values

t₂ = 3.47 s

now, time elapsed is = t₂ - t₁ = 3.47 - 0.0028 = 3.467 s

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3 years ago

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liq [111]

Answer:

x = 17.88[m]

Explanation:

We can find the components of the initial velocity:

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We have to remember that the acceleration of gravity will be worked with negative sign, since it acts in the opposite direction to the movement in direction and the projectile upwards.

g = - 9.81[m/s^2]

Now we must find the time it takes for the projectile to hit the ground, as the problem mentions that it does not impact on the board.

$y=y_{o} +(v_{y} )_{o} *t-0.5*g*(t)^{2} \\0=1.9+(8.81*t)-(4.905*t^{2})\\-1.9=8.81*t*(1-0.5567*t)\\t=0\\t=1.796[s]$

With this time we can calculate the horizontal distance:

$x=(v_{x})_{o} *t\\x=9.96*1.796\\x=17.88[m]$

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