
Concept explainers
The original “clock” used to define the length of the second was the daily rotation of Earth about its axis. Why has this clock been replaced by one based on the oscillation period of light waves emitted by atoms like cesium and rubidium?

The reason clocks based on the daily rotation has been replaced by the oscillation period of light waves emitted by atoms like cesium and rubidium.
Answer to Problem 1CA
The time period of the earth’s rotation is not fixed and to have a proper clock, it requires giving error free time period which is not possible. Hence, it is replaced by the oscillation period of light waves emitted by atoms like cesium and rubidium. The oscillation period of light waves emitted by atoms is fixed. Hence, the clock will give error free time.
Explanation of Solution
Introduction:
The earth’s rotation period is four minutes less than what we call a day. Due to this, the Sun’s position in the sky at noon is roughly fixed, but the stars slowly drift apart. So, the stars which are overhead today will slowly move in the sky until they will be overhead at noon 6 months apart.
If instead, a day was defined to be the time taken for earth to rotate, then according to above stated reason, the sun will be overhead on some day at midnight. The time period for oscillation of light waves emitted by the atoms of cesium and rubidium is always fixed. Hence, they will always give error free time period forever.
Conclusion:
The time given by the earth’s rotation won’t be same everyday as the earth’s rotation speed is not fixed. Hence, the clocks based on it will give wrong time. Hence, we use clocks based on the oscillations of light waves emitted by atoms like cesium and rubidium.
Want to see more full solutions like this?
Chapter 1 Solutions
Inquiry into Physics
- please help me solve this questions. show all calculations and a good graph too :)arrow_forwardWhat is the force (in N) on the 2.0 μC charge placed at the center of the square shown below? (Express your answer in vector form.) 5.0 με 4.0 με 2.0 με + 1.0 m 1.0 m -40 με 2.0 μCarrow_forwardWhat is the force (in N) on the 5.4 µC charge shown below? (Express your answer in vector form.) −3.1 µC5.4 µC9.2 µC6.4 µCarrow_forward
- An ideal gas in a sealed container starts out at a pressure of 8900 N/m2 and a volume of 5.7 m3. If the gas expands to a volume of 6.3 m3 while the pressure is held constant (still at 8900 N/m2), how much work is done by the gas? Give your answer as the number of Joules.arrow_forwardThe outside temperature is 25 °C. A heat engine operates in the environment (Tc = 25 °C) at 50% efficiency. How hot does it need to get the high temperature up to in Celsius?arrow_forwardGas is compressed in a cylinder creating 31 Joules of work on the gas during the isothermal process. How much heat flows from the gas into the cylinder in Joules?arrow_forward
- The heat engine gives 1100 Joules of energy of high temperature from the burning gasoline by exhausting 750 Joules to low-temperature . What is the efficiency of this heat engine in a percentage?arrow_forwardL₁ D₁ L₂ D2 Aluminum has a resistivity of p = 2.65 × 10 8 2. m. An aluminum wire is L = 2.00 m long and has a circular cross section that is not constant. The diameter of the wire is D₁ = 0.17 mm for a length of L₁ = 0.500 m and a diameter of D2 = 0.24 mm for the rest of the length. a) What is the resistance of this wire? R = Hint A potential difference of AV = 1.40 V is applied across the wire. b) What is the magnitude of the current density in the thin part of the wire? Hint J1 = c) What is the magnitude of the current density in the thick part of the wire? J₂ = d) What is the magnitude of the electric field in the thin part of the wire? E1 = Hint e) What is the magnitude of the electric field in the thick part of the wire? E2 =arrow_forwardplease helparrow_forward
- AstronomyPhysicsISBN:9781938168284Author:Andrew Fraknoi; David Morrison; Sidney C. WolffPublisher:OpenStaxPhysics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningPhysics for Scientists and Engineers, Technology ...PhysicsISBN:9781305116399Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
- University Physics Volume 3PhysicsISBN:9781938168185Author:William Moebs, Jeff SannyPublisher:OpenStaxUniversity Physics Volume 1PhysicsISBN:9781938168277Author:William Moebs, Samuel J. Ling, Jeff SannyPublisher:OpenStax - Rice UniversityPrinciples of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning





