Answer:
a The kinetic energy is 
b The height of the center of mass above that position is 
Explanation:
From the question we are told that
The length of the rod is 
The mass of the rod 
The angular speed at the lowest point is 
Generally moment of inertia of the rod about an axis that passes through its one end is
Substituting values
Generally the kinetic energy rod is mathematically represented as
From the law of conservation of energy
The kinetic energy of the rod during motion = The potential energy of the rod at the highest point
Therefore
To be referenced, it would be true
Answer:
(c) more than 500
Explanation:
Until 2019, more than 3000 planetary systems have been discovered that contain more than 4000 exoplanets, since some of these systems contain multiple planets. Most known extrasolar planets are gas giants equal to or more massive than the planet Jupiter, with orbits very close to its star.
<span>3598 seconds
The orbital period of a satellite is
u=GM
p = sqrt((4*pi/u)*a^3)
Where
p = period
u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits.
a = semi-major axis of orbit.
Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So
u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2
The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So
150000 m + 3.396x10^6 m = 3.546x10^6 m
Substitute the known values into the equation for the period. So
p = sqrt((4 * pi / u) * a^3)
p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3)
p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3)
p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3)
p = sqrt(1.2945785x10^7 s^2)
p = 3598.025212 s
Rounding to 4 significant figures, gives us 3598 seconds.</span>
For a star that has the same apparent brightness as Alpha Centauri A ( 2.7×10−8watt/m2 is mathematically given as
L=2.7*10^30w
<h3> What is its luminosity?</h3>
Generally, the equation for the luminosity is mathematically given as
L=4*\pi^2*b
Therefore
L=4*\pi^2*b
L=4* \pi *(2.83*10^{18})*2.7*10^{-8}
L=2.7*10^30
In conclusion, the luminosity
L=2.7*10^30w
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