Answer:
a) v = 2,9992 10⁸ m / s
, b) Eo = 375 V / m
, B = 1.25 10⁻⁶ T,
c) λ = 3,157 10⁻⁷ m, f = 9.50 10¹⁴ Hz
, T = 1.05 10⁻¹⁵ s
, UV
Explanation:
In this problem they give us the equation of the traveling wave
E = 375 cos [1.99 10⁷ x + 5.97 10¹⁵ t]
a) what the wave velocity
all waves must meet
v = λ f
In this case, because of an electromagnetic wave, the speed must be the speed of light.
k = 2π / λ
λ = 2π / k
λ = 2π / 1.99 10⁷
λ = 3,157 10⁻⁷ m
w = 2π f
f = w / 2 π
f = 5.97 10¹⁵ / 2π
f = 9.50 10¹⁴ Hz
the wave speed is
v = 3,157 10⁻⁷ 9.50 10¹⁴
v = 2,9992 10⁸ m / s
b) The electric field is
Eo = 375 V / m
to find the magnetic field we use
E / B = c
B = E / c
B = 375 / 2,9992 10⁸
B = 1.25 10⁻⁶ T
c) The period is
T = 1 / f
T = 1 / 9.50 10¹⁴
T = 1.05 10⁻¹⁵ s
the wavelength value is
λ = 3,157 10-7 m (109 nm / 1m) = 315.7 nm
this wavelength corresponds to the ultraviolet
Answer:
40 km/h
Explanation:
First...
Look at the formula speed is equal to the distance over time or s = d/t.
Next...
Use the formula: 240/6.0
Finally...
Solve: 40
So the answer: 40 km/h
As altitude increases, temperature increases.
The stratosphere is the part of the atmosphere that starts in the tropopause and ends in the estratopause. In the troposphere, the air is close to the Earth surface. The air surface can absorb more sunlight energy than the air, so the Earth surface heats the air. As you go higher, the distance to the Earth surface is higher, so the temperature is lower. The troposphere ends in the tropopause, where this trend changes. In the estratopause, there is a lot of ozone, which absorbs the dangerous UV radiation and converts into heat. That heat warms the air. So the air which is close to the estratopause is warm because of the heat released by the ozone reactions. The tropopause is far from the Earth surface and far from the ozone layer, that’s why it is cold. So the tropopause is cold and the estratopause is warm, which means: the air becomes warmer <span>as you rise above the tropopause until you get to the estratopause.</span>
Answer:
The velocity after 2 seconds can be found through:
V = u +a*t
Where V is final velocity, u is initial velocity, a is acceleration and t is time.
V = 0 + 2* 2= 4 meters/second
The distance (s) can be found through:
V^2= u^2 +2*a* s
Where V is final velocity, u is initial velocity, a is acceleration.
4^2= 0^2 + 2 *2*s
16= 0 + 4s
s= 4 meters
Distance (s) can also be found through:
s= ut + 1/2 at^2
s= 0+ 1/2 *2*2^2= 1 *2*2
s= 4 meters
Explanation:
