the Orbital Velocity is the velocity sufficient to cause a natural or artificial satellite to remain in orbit. Inertia of the moving body tends to make it move on in a straight line, while gravitational force tends to pull it down. The orbital path, elliptical or circular, representing a balance between gravity and inertia, and it follows a rue that states that the more massive the body at the centre of attraction is, the higher is the orbital velocity for a particular altitude or distance.
Answer:
stars will emit more light due to their Luminosity, so they look very bright.
Explanation:
Luminous refers to..,
- The total amount of energy radiated by a star or other celestial object per second.
- Therefore it is the power output of a star.
Most of the really bright stars in our sky are not that very close to us yet they look bright because of the Luminosity of the star.
These stars are intrinsically so luminous.
A star's power output across all wavelengths is called its bolometric luminosity.
A star with large luminosity will have more measure of radiated electromagnetic power meaning.
so it will emit more light than a low luminosity star.
Hence,
those stars can easily be seen even across great distance.
learn more about Luminosity of the star here:
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knives, axe ,blades and nails are examples of potential energy
or sharp objects
Answer:
The pressure drop predicted by Bernoulli's equation for a wind speed of 5 m/s
= 16.125 Pa
Explanation:
The Bernoulli's equation is essentially a law of conservation of energy.
It describes the change in pressure in relation to the changes in kinetic (velocity changes) and potential (elevation changes) energies.
For this question, we assume that the elevation changes are negligible; so, the Bernoulli's equation is reduced to a pressure change term and a change in kinetic energy term.
We also assume that the initial velocity of wind is 0 m/s.
This calculation is presented in the attached images to this solution.
Using the initial conditions of 0.645 Pa pressure drop and a wind speed of 1 m/s, we first calculate the density of our fluid; air.
The density is obtained to be 1.29 kg/m³.
Then, the second part of the question requires us to calculate the pressure drop for a wind speed of 5 m/s.
We then use the same formula, plugging in all the parameters, to calculate the pressure drop to be 16.125 Pa.
Hope this Helps!!!
