Numerical Solution of Algebraic Equations(Roots of Equations)4: The equation of motion of a mass is given as:y = (0.51)e 0. - (0.01)e0.2,where y is the displacement of the mass in meters and t is the time in seconds.Determine, numerically to three decimal-place accuracy, the time required forthe displacement to reduce to half its initial value.

The bisection method uses the midpoint of an interval to perform the iterations to approximate the…

Answered: The equation of motion of a mass is…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter5: Inverse, Exponential, And Logarithmic Functions

Section: Chapter Questions

Problem 73RE

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Van der Waals Equation In Exercise 18 at the end of Section 2.3, we discussed the ideal gas law, which shows the relationship among volume V, pressure p, and temperature T for a fixed amount 1 mole of a gas. But chemists believe that in many situations, the van der Waals equation gives more accurate results. If we measure temperature T in kelvins, volume V in liters, and pressure p in atmosphere 1 atm is the pressure exerted by the atmosphere at sea level, then the relationship for carbon dioxide is given by p=0.082TV0.0433.592V2atm What volume does this equation predict for 1 mole of carbon dioxide at 500 kelvins and 100 atm?Suggestion: Consider volumes ranging from 0.1 to 1 liter.
Radius of a Shock Wave An explosion produces a spherical shock wave whose radius R expands rapidly. The rate of expansion depends on the energy E of the explosion and the elapsed time t since the explosion. For many explosions, the relation is approximated closely by R=4.16E0.2t0.4. Here R is the radius in centimeters, E is the energy in ergs, and t is the elapsed time in seconds. The relation is valid only for very brief periods of time, perhaps a second or so in duration. a. An explosion of 50 pounds of TNT produces an energy of about 1015 ergs. See Figure 2.71. How long is required for the shock wave to reach a point 40 meters 4000 centimeters away? b. A nuclear explosion releases much more energy than conventional explosions. A small nuclear device of yield 1 kiloton releases approximately 91020 ergs. How long would it take for the shock wave from such an explosion to reach a point 40 meters away? c. The shock wave from a certain explosion reaches a point 50 meters away in 1.2 seconds. How much energy was released by the explosion? The values of E in parts a and b may help you set an appropriate window. Note: In 1947, the government released film of the first nuclear explosion in 1945, but the yield of the explosion remained classified. Sir Geoffrey Taylor used the film to determine the rate of expansion of the shock wave and so was able to publish a scientific paper concluding correctly that the yield was in the 20-kiloton range.

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The equation of motion of a mass is given as: y = (0.51)e 0.-(0.01)e0.2, where y is the displacement of the mass in meters and t is the time in seconds. Determine, numerically to three decimal-place accuracy, the time required for the displacement to reduce to half its initial value.