Chapter 12, Problem 62P

(a)

To determine

To calculate:The shock wave angle.

Answer to Problem 62P

The shock wave angle is $2.31°$ .

Explanation of Solution

Given:

Speed of meteorite $\left(v\right)$ $=8500\text{ m/s}$

Temperature $\left(T\right)$ $=20°\text{C}$

Formula used:

Shock wave is defined as, it is the strong pressure wave in an elastic medium like air, water or any solid and it is produced by supersonic speeds of an object.

Mathematically the shock wave is defined as:

  $\sin \theta =\frac{1}{M}\text{ or }\frac{C}{V}$

Where,

  $C=$ Velocity of sound in air at $20°\text{C}$ = 343 m/s,

  $V=$ Velocity of an object.

Calculation:

Substitute the values in the equation,

  $\sin \theta =\frac{343\text{ m/s}}{8500\text{ m/s}}$

  $\theta ={\sin }^{-1}(0.04)$

  $\theta =2.31°$

Conclusion:

The shock wave angle is $2.31°$ , that produces in the air just before entering the ocean.

(b)

To determine

To calculate:The shock wave angle.

Answer to Problem 62P

The shock wave angle $\theta =10.3°$ .

Explanation of Solution

Given:

Speed of meteorite $=8500\text{ m/s}$ .

Temperature $\left(T\right)$ $=20°\text{C}$

Formula used:

The sine of the angle is equal to the ratio of velocity of sound to the velocity of an object is known as shock wave angle.

Mathematically,

  $\sin \theta =\frac{C}{V}\text{ or }\frac{1}{M}$

  $C=$ Velocity of sound in air at $20°\text{C}$ = 343 m/s,

  $V=$ Velocity of an object.

Calculation:

Substitute the values in the equation,

  $\sin \theta =\frac{\text{1522 m/s}}{8500\text{ m/s}}$

  $\theta ={\sin }^{-1}(0.17)$

  $\theta =10.3°$

Conclusion:

The shock wave angle, that produces in the water just after entering the ocean is $10.3°$ .