-- The source of most of the energy that radiates from the sun is nuclear energy.
-- Most of the energy that radiates from the sun is electromagnetic energy.
-- Heat energy is part of the electromagnetic energy that radiates from the sun.
Other parts include radio, microwave, visible light, ultraviolet, and X-ray energy.
Answer:
I am sure the awnser to this is
B. vibrate
Answer:
7) λ = 0.5 m, 8) f = 4.8 10¹⁴ Hz
Explanation:
The speed of an electromagnetic wave is
c = λ f
where c is the speed of light in vacuum c = 3 10⁸ m / s
7) indicate the frequency f = 6.0 10⁸ Hz
we do not know the wavelength
λ = c / f
we calculate
λ = 3 10⁸ / 6.0 10⁸
λ = 0.5 m
8) indicate the wavelength λ = 6.25 10-7 m
we do not know the frequency
f = c / λ
we calculate
f = 3 10⁸ / 6.25 10⁻⁷
f = 0.48 10¹⁵ Hz
f = 4.8 10¹⁴ Hz
Answer:
newton's first law hope this helped
Answer:
<em>t=14.96 sec</em>
Explanation:
<u>Diagonal Launch
</u>
It's a physical event that happens where an object is thrown in free air (no friction) forming an angle with the horizontal reference. The object then describes a path called a parabola.
The object will reach its maximum height and then return to the height from which it was launched. The equation for the height is
:

Where vo is the initial speed,
is the angle, t is the time and g is the acceleration of gravity
.
In this problem we'll assume the arrow was launched from the ground level (won't consider the archer's height). Thus
, and:

The value of y is zero twice: when t=0 (at launching time) and in t=
when it goes back to the ground. We need to find that time
by making

Dividing by 

Then we find the total flight time as

