/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 About 50,000 years ago, a meteor... [FREE SOLUTION] | 91影视

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About 50,000 years ago, a meteor crashed into the earth near present-day Flagstaff, Arizona. Measurements from 2005 estimate that this meteor had a mass of about 1.4 \(\times\) 10\(^8\) kg (around 150,000 tons) and hit the ground at a speed of 12 km/s. (a) How much kinetic energy did this meteor deliver to the ground? (b) How does this energy compare to the energy released by a 1.0-megaton nuclear bomb? (A megaton bomb releases the same amount of energy as a million tons of TNT, and 1.0 ton of TNT releases 4.184 \(\times\) 10\(^9\) J of energy.)

Short Answer

Expert verified
The meteor delivered 1.008 脳 10鹿鈦 J of energy, about 24% of a 1-megaton bomb's energy.

Step by step solution

01

Identify Required Formula

To find the kinetic energy (KE) delivered by the meteor, we use the formula for kinetic energy:\[KE = \frac{1}{2} mv^2\]where \(m\) is the mass of the meteor, and \(v\) is its velocity.
02

Convert Velocity

Before plugging the values into the kinetic energy formula, convert the velocity from kilometers per second to meters per second:\[12 \text{ km/s} = 12,000 \text{ m/s}\]
03

Substitute Values into Kinetic Energy Formula

Now, substitute the mass \(m = 1.4 \times 10^8\) kg and the velocity \(v = 12,000\) m/s into the kinetic energy formula:\[KE = \frac{1}{2} \times 1.4 \times 10^8 \times (12,000)^2\]Calculate the result to find the kinetic energy.
04

Calculate Kinetic Energy

Carrying out the calculation:\[KE = 0.5 \times 1.4 \times 10^8 \times 144,000,000 = 1.008 \times 10^{15} \text{ J}\]
05

Energy of a 1-Megaton Nuclear Bomb

Calculate the energy released by a 1.0-megaton nuclear bomb. One megaton is one million tons, and 1 ton of TNT releases 4.184 \(\times\) 10\(^9\) J of energy:\[1 \text{ megaton} = 1,000,000 \times 4.184 \times 10^9 \text{ J} = 4.184 \times 10^{15} \text{ J}\]
06

Compare Kinetic Energy to Nuclear Bomb

Compare the meteor's kinetic energy to the energy of the 1-megaton bomb:- Meteor's KE = 1.008 \(\times\) 10\(^15\) J- 1-Megaton Bomb = 4.184 \(\times\) 10\(^15\) JThe meteor's kinetic energy was approximately \(\frac{1.008}{4.184} \approx 0.241\) or about 24% of that of a 1-megaton nuclear bomb.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Meteor Impact
Imagine a gigantic rock from space hurtling towards Earth, racing at an astonishing speed. This is exactly what happened near Flagstaff, Arizona 50,000 years ago when a meteor struck the planet.
The meteor had a massive weight of about 150,000 tons, which is equivalent to 1.4 \( \times \) 10\(^8\) kg.
While zooming through space, it reached a speed of 12 kilometers per second. This high rate of motion translates to 12,000 meters per second.
What makes this event so fascinating? It's the energy with which the meteor hit the Earth. To understand this, we turn to kinetic energy, a key concept that calculates the energy an object possesses due to its motion. The formula used for calculating kinetic energy is:
  • \( KE = \frac{1}{2}mv^2 \)
where \( m \) represents mass and \( v \) represents velocity. Substituting the meteor's mass and velocity into this formula reveals that the energy delivered upon impact was 1.008 \( \times \) 10\(^ { 15 }\) J.The crater formed from this impact provides physical evidence of the tremendous force with which meteors can strike the Earth, altering landscapes and ecosystems.
Energy Comparison
Kinetic energy can be compared to other forms of energy, giving us a way to relate one tremendous force to another familiar one. In this case, we compare the meteor's kinetic energy to the energy released by a 1-megaton nuclear bomb.
A nuclear bomb of this size releases energy equivalent to a million tons of TNT. If we delve into the energy in one ton of TNT, we discover it emits 4.184 \( \times \) 10\(^ 9 \) Joules. So, a 1-megaton bomb releases this much energy when we multiply:
  • 1,000,000 (million tons) \( \times \) 4.184 \( \times \) 10\(^ 9 \) J = 4.184 \( \times \) 10\(^ { 15 }\) J
Now, how do these energies compare? The meteor's energy was quite significant, but it amounted to only approximately 24% of the energy released by a megaton nuclear bomb. This comparison helps us visualize the massive energy released during both events. It underscores the meteor's incredible speed and mass, even if it didn鈥檛 match the bomb鈥檚 ultimate output.
Nuclear Bomb Energy
Nuclear bomb energy is a measurement often used because of its vast potential to release energy in explosions.
The energy comes from nuclear reactions, either fission or fusion, which can release a tremendous amount of energy from tiny amounts of material. This energy potential is so vast that nuclear bombs are often described by their explosive power in "megaton" equivalents of TNT.
A 1-megaton nuclear bomb, for example, releases as much energy as one million tons of TNT. This massive release of energy can cause widespread destruction, not only by the blast itself but also by the heat generated and radiation released.
Nuclear bomb energy helps us understand the scale of dynamic cosmic events, like meteor impacts, by providing a familiar reference point.
It gives us a chilling yet fascinating context to compare the powerful energy unleashed by forces of nature, prompting us to respect both the engineered and natural potentials around us.

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Most popular questions from this chapter

A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from \(x = 0\) to \(x = 6.9\) m as you apply a force with \(x\)-component \(F_x = - [20.0 \, \mathrm{N} + (3.0 \, \mathrm{N/m})x]\). How much work does the force you apply do on the cow during this displacement?

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An airplane in flight is subject to an air resistance force proportional to the square of its speed v. But there is an additional resistive force because the airplane has wings. Air flowing over the wings is pushed down and slightly forward, so from Newton's third law the air exerts a force on the wings and airplane that is up and slightly backward (\(\textbf{Fig. P6.94}\)). The upward force is the lift force that keeps the airplane aloft, and the backward force is called \(induced \, drag\). At flying speeds, induced drag is inversely proportional to \(v^2\), so the total air resistance force can be expressed by \(F_air = \alpha v^{2} + \beta /v{^2}\), where \(\alpha\) and \(\beta\) are positive constants that depend on the shape and size of the airplane and the density of the air. For a Cessna 150, a small single-engine airplane, \(\alpha = 0.30 \, \mathrm{N} \cdot \mathrm{s^{2}/m^{2}}\) and \(\beta = 3.5 \times 10^5 \, \mathrm{N} \cdot \mathrm{m^2/s^2}\). In steady flight, the engine must provide a forward force that exactly balances the air resistance force. (a) Calculate the speed (in km/h) at which this airplane will have the maximum \(range\) (that is, travel the greatest distance) for a given quantity of fuel. (b) Calculate the speed (in km/h) for which the airplane will have the maximum \(endurance\)(that is, remain in the air the longest time).

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