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2 years ago

9

A fighter plane is flying overhead at mach 1.20. What angle does the wave front of the shock wave produced make relative to the

plane's direction of motion (in degrees)?

1 answer:

Sever21 [200]2 years ago

7 0

Answer: 56.44°

Explanation:

<u>Given:</u>

Let u represent the current speed of the plane, <u>1.2 Mach</u>

<em>Converting to SI Units (m/s):</em>

= (1.2 mach)(340 ms^-1 / 1 Mach)

u = 408 m/s

Speed of sound in air, v = 340 m/s

<u>Find:</u>

Angle the wave front of the shock wave relative to the plane's direction of motion, θ

We have, sinθ = speed of sound / speed of object

sinθ = v / u

θ = sin^-1 (v / u)

= sin^-1 (340 / 408)

θ = 56.44°

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3 years ago

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From the question, Abby is standing 5.00m in front of one of the speakers, perpendicular to the line joining the speakers.

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Hence, the path difference is

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Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

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Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
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<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

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$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

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3 years ago