Answer: Highest point from the ground that ball reaches = 5.102
Explanation:
Formula -
v² - u² = 2as
final velocity v = 0
initial velocity u = 10m/s
acceleration a = g (gravity)=9.8
distance s = 3m+x (x is the height after 3m that object reaches)
0-(10)² = 2 (-g)*(3+x)
-100 = -2g*(3=x)
(3+x)=100/2g
3+x = 100/2*9.8 = 100/19.6 = 5.102
x=5.102-3
x=2.102
So, the highest point will be
3+x = 3+2.102
=5.102m
Accelerating objects are objects which are changing their velocity - either the speed (i.e., magnitude of the velocity vector) or the direction. An object undergoing uniform circular motion is moving with a constant speed. ... The net force acting upon such an object is directed towards the center of the circle.
The electromagnetic spectrum is divided into seven different frequency ranges, which are, from lowest to highest, radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays.
The electromagnetic spectrum includes electromagnetic waves with frequencies between one hertz and above 10²⁵ hertz, or wavelengths between thousands of kilometers and a small portion of the size of an atomic nucleus. The electromagnetic waves found within each of these bands are referred to by a different name; starting at the low frequency (long wavelength) end of the spectrum, these are radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. This frequency range is divided into separate bands.
To know more about electromagnetic spectrum
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Answer: The distance between particles, and the amount of electric charge they carry.
Explanation:
Charles Coulomb wanted to figure out the strength of the force between two objects and these were the two most independent factors.
Answer: Looked it up but
Explanation:
When the skater lands on the track, the vertical component of his kinetic energy is converted to thermal energy. You can do experiments where there is no loss to thermal energy (only PE and KE conversions) by turning friction off and by making sure the skater doesn't leave the track.