Numerical Solution of Algebraic Equations(Roots of Equations)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 Newton-Raphson method uses the formula: tn+1=tn-f(tn)f'(tn) to perform the iterations and to…

Answered: The equation of motion of a mass is…

College Algebra

10th Edition

ISBN:9781337282291

Author:Ron Larson

Publisher:Ron Larson

Chapter1: Equations, Inequalities, And Mathematical Modeling

Section1.2: Linear Equations In One Variable

Problem 79E

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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)e.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.