Answer:
When the temperature decreases the particals start to slow down.
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:
I am pretty sure it is B.
Explanation:
I hope this helped if it didn't I am truly sorry
Either one is fun and great to play!
Answer: 405.3 minutes
Explanation: In order to explain this problem we have to use the following:
Fisrtly we calculate the volume of the wire, this is given by:
Vwire=π*r^2*L where r and L are the radius and L the length of teh wire, respectively.
Vwire=π*1.25*10^-3*0.26=1.27*10^-6 m^3
then the number of the total electrons in tthe wire volume is given by;
n° electrons in the wire=ρ*Vwire=8.4*10^28*1.27*10^-6 m^3=1.07 *10^23
Finally, considering the current in the wire equal to 4.4*10^18 electrons/s
the time consuming to extract all the electrons from the wire is given by:
t= total electrons in the wire/ current=1.067*10^23/4.4*10^18=24,318 s
equivalent to 405.3 minutes
