Answer:
neutrons protons electrons
Explanation:
after the Big Bang, the universe was filled with neutrons, protons, electrons, anti-electrons, photons and neutrinos.
The acceleration of gravity on Earth is 9.8 m/s² .
The speed of a falling object keeps increasing smoothly,
in such a way that the speed is always 9.8 m/s faster than
it was one second earlier.
If you 'drop' the penny, then it starts out with zero speed.
If you also start the clock at the same instant, then
After 1.10 sec, Speed = (1.10 x 9.8) = 10.78 meters/sec
After 1.85 sec, Speed = (1.85 x 9.8) = 18.13 meters/sec
But you want this second one given in a different unit of speed.
OK then:
= (18.13 meter/sec) x (3,600 sec/hr) x (1 mile/1609.344 meter)
= (18.13 x 3,600 / 1609.344) (mile/hr) = 40.56 mph (rounded)
We did notice that in an apparent effort to make the question
sound more erudite and sophisticated, you decided to phrase
it in terms of 'velocity'. We can answer it in those terms, if we
ASSUME that there is no wind, and the penny therefore doesn't
acquire any horizontal component of motion on its way down.
With that assumption in force, we are able to state unequivocally
and without fear of contradiction that each 'speed' described above ...
with the word 'downward' appended to it ... does become a 'velocity'.
Answer:
Because the wavelengths of macroscopic objects are too short for them to be detectable.
Explanation:
Wavelength of an object is given by de Broglie wavelength as:
Where, 'h' is Planck's constant, 'm' is mass of object and 'v' is its velocity.
So, for macroscopic objects, the mass is very large compared to microscopic objects. As we can observe from the above formula, there is an inverse relationship between the mass and wavelength of the object.
So, for vary larger masses, the wavelength would be too short and one will find it undetectable. Therefore, we don't observe wave properties in macroscopic objects.
The Hertzsprung- Russel diagram is used to how the relationship between the absolute magnitudes of stars and their effective temperatures.
