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Chapter 9: Problem 86

A 2 -mm-diameter meteor of specific gravity 2.9 has a speed of \(6 \mathrm{km} / \mathrm{s}\) at an altitude of \(50,000 \mathrm{m}\) where the air density is \(1.03 \times\) \(10^{-3} \mathrm{kg} / \mathrm{m}^{3} .\) If the drag coefficient at this large Mach number condition is \(1.5,\) determine the deceleration of the meteor

Short Answer

Expert verified
The deceleration of the meteor is calculated by following the four steps of firstly finding the radius, volume, and mass of the meteor, then determining the area, calculating the drag force, and finally using these results in Newton's second law of motion to find the deceleration. The exact value of the deceleration will depend on the results of these calculations.

Step by step solution

01

Calculate the Radius, Volume, and Mass of the Meteor

The given diameter is 2 mm or 0.002 m, so the radius \( r \) is half of this value, or 0.001 m. The volume \( V \) is given by the formula \( 4/3 πr^3 \). The mass \( m \) can be found by multiplying the volume by the specific gravity, and then by the density of water (since specific gravity is relative to water), which is 1000 kg/m³. So, \( m = V × SG × density\_of\_water \).
02

Calculate the Area

The area \( A \) over which the drag force acts is that of a circle \( πr^2 \).
03

Calculate the Drag Force

The drag force \( F_d \) is found using the formula \( 0.5 × air\_density × velocity^2 × drag\_coefficient × A \). The given air density is \(1.03 × 10^{-3} kg/m³ \), the velocity is 6,000 m/s (converted from 6 km/s), and the drag coefficient is 1.5.
04

Determine Deceleration

The deceleration \( a \) is found using Newton’s second law of motion \( F = ma \) rearranged as \( a = F/m \). The net force acting on the meteor is the drag force, as it is the only force acting in the opposite direction to motion. Calculate the deceleration by dividing the drag force by the mass of the meteor.

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Most popular questions from this chapter

Compare the rise velocity of an \(\frac{1}{8}\) -in.-diameter air bubble in water to the fall velocity of an \(\frac{1}{8}\) -in.- -diameter water drop in air. Assume each to behave as a solid sphere.

An airplane flies at \(150 \mathrm{km} / \mathrm{hr}\). (a) The airplane is towing a banner that is \(b=0.8 \mathrm{m}\) tall and \(\ell=25 \mathrm{m}\) long. If the drag coefficient based on area \(b \ell\) is \(C_{D}=0.06,\) estimate the power required to tow the banner. (b) For comparison, determine the power required if the airplane was instead able to tow a rigid flat plate of the same size. (c) Explain why one had a larger power requirement (and larger drag) than the other. (d) Finally, determine the power required if the airplane was towing a smooth spherical balloon with a diameter of \(2 \mathrm{m}\)

On a day without any wind, your car consumes \(x\) gallons of gasoline when you drive at a constant speed, \(U\), from point \(A\) to point \(B\) and back to point \(A\). Assume that you repeat the journey. driving at the same speed, on another day when there is a steady wind blowing from \(B\) to \(A\). Would you expect your fuel consumption to be less than, equal to, or greater than \(x\) gallons for this windy round-trip? Support your answer with appropriate analysis.

A nonspinning ball having a mass of 3 oz. is thrown vertically upward with a velocity of 100 mph and has zero velocity at a height 250 ft above the release point. Assume that the air drag on the ball is constant and find this constant "average" air drag. Neglect the buoyant force of air on the ball.

(See The Wide World of Fluids article "At \(12,600 \mathrm{mpg}\) It Doesn't cost Much to 'Fill 'er Up," section 9.3.3.) (a) Determine the power it takes to overcome aerodynamic drag on a small \(\left(6 \mathrm{ft}^{2}\right.\) cross section), streamlined \(\left(C_{D}=0.12\right)\) vehicle traveling \(15 \mathrm{mph}\) (b) Compare the power calculated in part (a) with that for a large \(\left(36 \mathrm{ft}^{2} \text { cross-sectional area }\right),\) nonstreamlined \(\left(C_{D}=0.48\right) \mathrm{SUV}\) traveling 65 mph on the interstate.
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