An object is moving with the speed of light around the Earth. How much time will it take to complete one round trip along the eq
1 answer:
Answer:
0.14 seconds
Explanation:
The speed of light in vacuum is approximately 3.0*10^8. The distance that would be covered by the object would be equivalent to the circumference of the cross-section of the earth on the equator.
Circumference = 2*6400000 =4.02*10^7
Time = distance/speed = 4.2*10^7 / 3.0*10^8 =0.14s
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Answer:
(a) f = 0.58Hz
(b) vmax = 0.364m/s
(c) amax = 1.32m/s^2
(d) E = 0.1J
(e) x(t)=0.1m*cos(2π(0.58s^{-1})t)
Explanation:
(a) The frequency of the oscillation, in a spring-mass system, is calculated by using the following formula:
(1)
k: spring constant = 20.0N/m
m: mass = 1.5kg
you replace the values of m and k for getting f:
The frequency of the oscillation is 0.58Hz
(b) The maximum speed is given by the following relation:
(2)
A: amplitude of the oscillations = 10.0cm = 0.10m
The maximum speed of the mass is 0.364 m/s.
The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.
(c) The maximum acceleration is given by the following formula:
The maximum acceleration is 1.32 m/s^2
The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.
(d) The total energy of the system is calculated with the maximum potential elastic energy:
The total energy is 0.1J
(e) The displacement as a function of time is: