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FrozenT [24]
1 year ago
10

A large meteoroid enters the Earth's atmosphere at a speed of 20.0 km/s and is not significantly slowed before entering the ocea

n. (a) What is the Mach angle of the shock wave from the meteoroid in the lower atmosphere?
Physics
1 answer:
Shalnov [3]1 year ago
5 0

The shock wave from the meteoroid in the lower atmosphere has a Mach angle of 0.948°.

(a) The meteoroid's speed v_s=20 \mathrm{~km} / \mathrm{s}

$=20 \times 10^3 \mathrm{~m} / \mathrm{s}$

Air sound wave speed &v=331 \mathrm{~m} / \mathrm{s} \\

Speed of the shock wave in Mach &\qquad \theta=\sin ^{-1}\left(\frac{v}{v_s}\right)

                                                            $$\begin{aligned}&=\sin ^{-1}\left(\frac{331 \mathrm{~m} / \mathrm{s}}{20 \times 10^3 \mathrm{~m} / \mathrm{s}}\right) \\&=0.948^{\circ}\end{aligned}$$

Hence, 0.948° is the Mach angle of the shock wave from the meteoroid in the lower atmosphere.

<h3>What is the speed of the meteoroid?</h3>

A meteoroid's speed can be loosely broken down into three categories: slow, medium, and fast.

  • Slow meteors move around the sun at a leisurely pace of about 32 kilometers per second (20 miles per second).
  • Medium-speed meteors travel around the sun at approximately 50 kilometers per second (30 miles per second),
  • while fast meteors zoom past at over 120 kilometers per second (75 miles per second)!

To learn more about meteoroid, visit:

brainly.com/question/1939309

#SPJ4

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1 year ago
A 1900 kg car rounds a curve of 55 m banked at an angle of 11 degrees? . If the car is traveling at 98 km/h, How much friction f
Nat2105 [25]

Answer:

22000 N

Explanation:

Convert velocity to SI units:

98 km/h × (1000 m / km) × (1 h / 3600 s) = 27.2 m/s

Draw a free body diagram.  There are three forces acting on the car.  Normal force perpendicular to the bank, gravity downwards, and friction parallel to the bank.

I'm going to assume the friction force is pointed down the bank.  If I get a negative answer, that'll just mean it's actually pointed up the bank.

Sum of the forces in the radial direction (+x):

∑F = ma

N sin θ + F cos θ = m v² / r

Sum of the forces in the y direction:

∑F = ma

N cos θ - F sin θ - W = 0

To solve the system of equations for F, first solve for N and substitute.

N = (W + F sin θ) / cos θ

Substituting:

((W + F sin θ) / cos θ) sin θ + F cos θ = m v² / r

(W + F sin θ) tan θ + F cos θ = m v² / r

W tan θ + F sin θ tan θ + F cos θ = m v² / r

W tan θ + F (sin θ tan θ + cos θ) = m v² / r

W tan θ + F sec θ = m v² / r

F sec θ = m v² / r - W tan θ

F = m v² cos θ / r - W sin θ

F = m (v² cos θ / r - g sin θ)

Given that m = 1900 kg, θ = 11°, v = 27.2 m/s, and r = 55 m:

F = 1900 ((27.2)² cos 11 / 55 - 9.8 sin 11)

F = 21577 N

Rounding to two sig-figs, you need at least 22000 N of friction force.

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