/*! 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 33 (II) How much energy can be obta... [FREE SOLUTION] | 91Ó°ÊÓ

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(II) How much energy can be obtained from conversion of 1.0 gram of mass? How much mass could this energy raise to a height of 1.0 km above the Earth's surface?

Short Answer

Expert verified
1.0 gram of mass converts to \( 9.0 \times 10^{13} \) joules, enough to raise \( 9.18 \times 10^9 \) kg by 1 km.

Step by step solution

01

Determine the energy from mass conversion

Use Einstein's mass-energy equivalence formula to find the energy obtained from a mass of 1.0 gram. The formula is given by:\[ E = mc^2 \]where:- \( m \) is the mass in kilograms (1.0 gram = 0.001 kg),- \( c \) is the speed of light \( \approx 3.0 \times 10^8 \) meters per second.Substitute the values into the equation:\[ E = 0.001 \times (3.0 \times 10^8)^2 = 0.001 \times 9.0 \times 10^{16} = 9.0 \times 10^{13} \text{ joules} \]
02

Calculate the mass that can be raised by this energy

Use the work-energy principle, which states that the work done (energy used) to raise a mass is equal to \( mgh \), where:- \( m \) is the mass to be raised,- \( g \) is the acceleration due to gravity \( \approx 9.8 \text{ m/s}^2 \),- \( h \) is the height \( = 1000 \text{ m} \) or 1.0 km.Set the energy obtained from mass conversion equal to the potential energy:\[ E = mgh \]\[ 9.0 \times 10^{13} = m \times 9.8 \times 1000 \]Solve for \( m \):\[ m = \frac{9.0 \times 10^{13}}{9.8 \times 1000} \approx 9.18 \times 10^{9} \text{ kg} \]
03

Conclusion

The energy obtained from converting 1.0 gram of mass is enough to raise approximately \( 9.18 \times 10^{9} \) kilograms to a height of 1.0 kilometer above the Earth's surface.

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Key Concepts

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

Einstein's Theory
Albert Einstein revolutionized our understanding of physics with his mass-energy equivalence principle. This concept is beautifully encapsulated in the equation \[ E = mc^2 \] where:
  • \( E \) stands for energy,
  • \( m \) represents mass, and
  • \( c \) is the speed of light, approximately \( 3.0 \times 10^8 \) meters per second.

This means that a relatively small amount of mass can be converted into a tremendous amount of energy. For instance, converting just 1.0 gram of mass results in a whopping \( 9.0 \times 10^{13} \) joules of energy.

The equivalence of mass and energy implies that mass itself is a form of stored energy. This principle is a fundamental aspect of nuclear reactions, where large amounts of energy are released by converting small amounts of mass. It indeed forms the bedrock of modern physics and cosmology, transforming how we perceive matter and energy in the universe.
Potential Energy
Potential energy is a type of mechanical energy related to the position or configuration of an object. When we talk about potential energy in the context of gravitational fields, it is the energy stored by an object due to its position relative to the Earth.

Formula for gravitational potential energy is:\[ PE = mgh \]where:
  • \( m \) is the mass of the object,
  • \( g \) is the acceleration due to gravity, approximately \( 9.8 \text{ m/s}^2 \), and
  • \( h \) is the height above the reference point (in this case, the Earth's surface).

Take our exercise, for example; if we use the energy from converting 1.0 gram of mass, we can raise a huge 9.18 billion kilograms of mass to 1.0 kilometer height. This highlights the incredible energy-mass relationship and how it can be applied to do work, such as raising objects within a gravitational field.
Work-Energy Principle
The work-energy principle is crucial for understanding how forces affect motion. It states that work done by all forces acting on an object will result in a change in energy of that object. In mathematical terms, this can be represented as:\[ W = \DeltaKE + \DeltaPE \]where:
  • \( W \) is the work done on the object,
  • \( \Delta KE \) is the change in kinetic energy, and
  • \( \Delta PE \) is the change in potential energy.

In our given problem, this principle allows us to calculate the maximum mass that can be lifted by the energy derived from mass conversion. Essentially, we've equated the energy from mass conversion, \( 9.0 \times 10^{13} \) joules, with gravitational potential energy \( mgh \) to find out how much mass \( m \) can be potentially lifted.

This showcases the seamless transition from energy provided by conversion (in line with Einstein’s idea) to doing physical work like lifting mass, thereby interconnecting these magnificent concepts in a practical and understandable way.

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Most popular questions from this chapter

An astronaut on a spaceship traveling at 0.75\(c\) relative to Earth measures his ship to be 23 m long. On the ship, he eats his lunch in 28 min. (\(a\)) What length is the spaceship according to observers on Earth? (\(b\)) How long does the astronaut's lunch take to eat according to observers on Earth?

The fictional starship \(Enterprise\) obtains its power by combining matter and antimatter, achieving complete conversion of mass into energy. If the mass of the \(Enterprise\) is approximately 6 \(\times 10^9\) kg. how much mass must be converted into kinetic energy to accelerate it from rest to one-tenth the speed of light?

(I) When a uranium nucleus at rest breaks apart in the process known as \(fission\) in a nuclear reactor, the resulting fragments have a total kinetic energy of about 200 \(M\)e\(V\). How much mass was lost in the process?

(II) A friend speeds by you in her spacecraft at a speed of 0.720\(c\). It is measured in your frame to be 4.80 m long and 1.35 m high. (\(a\)) What will be its length and height at rest? (\(b\)) How many seconds elapsed on your friend's watch when 20.0 s passed on yours? (\(c\)) How fast did you appear to be traveling according to your friend? (d) How many seconds elapsed on your watch when she saw 20.0 s pass on hers?

(II) An unstable particle is at rest and suddenly decays into two fragments. No external forces act on the particle or its fragments. One of the fragments has a speed of 0.60\(c\) and a mass of 6.68 \(\times 10^{-27}\) kg, while the other has a mass of 6.67 \(\times 10^{-27}\) kg, What is the speed of the less massive fragment?

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