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2 years ago

13

What is the difference between a sonic boom and a shock wave?

1 answer:

WINSTONCH [101]2 years ago

3 0

Answer:

When a sound source moves faster than the speed of sound, a shock wave is produced as the sound waves interfere. A sonic boom is the intense sound that occurs as the shock wave moves along the ground.

More importantly, all of the energy gets concentrated into a very small distance this is called a shock wave. In this case, the observer does not hear the approaching source at all until the shock wave hits with all of the energy in the wave. For sound waves, this can cause a very loud noise, called a sonic boom.

Sonic booms produced by aircraft flying supersonic at altitudes of less than 100 feet, creating between 20 and 144 pounds overpressure, have been experienced by humans without injury. Damage to eardrums can be expected when overpres- sures reach 720 pounds

Explanation:

marke as brainliest

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What is a covalent bond considered

vekshin1

A covalent bond is considered covalent

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3 years ago

A metallic wire has a diameter of 4.12mm. When the current in the wire is 8.00A, the drift velocity is 5.40×10−5m/s.What is the

podryga [215]

Answer:

$6.9\times 10^{28}m^{-3}$

Explanation:

We are given that

Diameter of wire=d=4.12 mm

Radius of wire=r $r=\frac{d}{2}=\frac{4.12}{2}=2.06mm=2.06\times 10^{-3} m$

$1mm=10^{-3} m$

Current=I=8 A

Drift velocity= $v_d=5.4\times 10^{-5} m/s$

We have to find the density of free electrons in the metal

We know that

Density of electron= $n=\frac{I}{v_deA}$

Using the formula

Density of free electrons= $\frac{8}{5.4\times 10^{-5}\times 1.6\times 10^{-19}\times 3.14\times (2.06\times 10^{-3})^2}$

By using Area of wire= $\pi r^2$

$\pi=3.14\\e=1.6\times 10^{-19} C$

Density of free electrons= $6.9\times 10^{28}m^{-3}$

3 0

3 years ago

Read 2 more answers

For a particular casting setup, the top of the sprue has a diameter of 0.030 m, and its length is 0.200 m. The volume flow rate

faust18 [17]

Answer with Explanation:

We are given that

Diameter=0.030 m

Length of sprue= $h_1$ =0.200 m

Metal volume flow rate,Q=0.03 $m^3/min$

Q= $\frac{0.03}{60}=5\times 10^{-4}m^3/s$ because 1 minute=60 seconds

Let 1 for the top and 2 for the bottom

$d_=0.030 m$

$h_2=0$

$A_1=\frac{\pi d^2}{4}=\frac{3.14\times (0.030)^2}{4}$

$A_1=7.065\times 10^{-4} m^2$

$v_1=\frac{Q}{A_1}=\frac{5\times 10^{-4}}{7.065\times 10^{-4}}$

$v_1=0.708 m/s$

Pressure at the top and bottom of the sprue is atmospheric

$h_1+\frac{v^2_1}{2g}=h_2+\frac{v^2_2}{2g}$

Substitute the values

$0.2+\frac{(0.708)^2}{2\cdot 9.8}=0+\frac[v^2_2}{2\cdot 9.8}$

$v^2_2=2\cdot 9.8\cdot \frac{0.2\cdot 9.8\cdot 2+0.501264}{2\cdot 9.8}=4.421264$

$v_2=\sqrt{4.421264}=2.1 m/s$

$Q=A_2v_2$

$5\times 10^{-4}=A_2\times 2.1$

$A_2=\frac{5\times 10^{-4}}{2.1}=2.381\times 10^{-4} m^2$

Reynolds number= $\frac{v_2D\rho}{\eta}$

$\eta=0.004 N.s/m^2$

$\rho=2700 kg/m^3$

Substitute the values then we get

Reynolds number= $\frac{2.1\times 0.03\times 2700}{0.004}$

Reynolds number=42525

The Reynolds number is greater than 4000 .Therefore, the flow is turbulent.

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3 years ago

How far away is the sun from earth in AU?

julia-pushkina [17]

That distance is the definition of 1 AU.

5 0

3 years ago

A 5-kg fish swimming at a speed of 1 m/s swallows an absent-minded 1-kg fish at rest. The speed of the larger fish after lunch i

8090 [49]

Answer:

Explanation:

So:

(

5

⋅

1

)

+

(

1

⋅

−

4

)

=

(

5

+

1

)

v

5

−

4

=

6

v

v

=

1

6

=

0.17

m

s

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2 years ago