As this happens over twelve seconds, you would take the total difference in velocities and divide it by twelve to find the change per second
44.0 m/s - 2.0 m/s = 42.0 m/s
42.0 m/s / 12 s = 3.5 m/s2
the acceleration of the rock would be 3.5 m/s2
Answer:
The answer is "Option C".
Explanation:
Computation modeling is used as the software for math, physics, or software engineering in the simulation and study of complicated processes. The software framework incorporates many parameters, that characterize the model under study. It is a computer program, in which the mathematical formula is generated by computers and research complex systems.
To solve the problem, use Kepler's 3rd law :
T² = 4π²r³ / GM
Solved for r :
r = [GMT² / 4π²]⅓
but first covert 6.00 years to seconds :
6.00years = 6.00years(365days/year)(24.0hours/day)(6...
= 1.89 x 10^8s
The radius of the orbit then is :
r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.89 x 10^8s)² / 4π²]⅓
= 6.23 x 10^11m
Well im not sure if this is the correct dating materials but here are some examples of Fundamentals of radiometric dating<span>Radioactive decay.
Accuracy of radiometric dating.
Closure temperature.
The age equation.
Uranium–lead dating method.
Samarium–neodymium dating method.
Potassium–argon dating method.
<span>Rubidium–strontium dating method.</span></span>