The correct answer is
<span>B) UV waves are higher frequency and carry more energy.
In fact, UV waves have higher frequency than visible light. For comparison, visible light has frequency in the range 430-770 THz (</span>
)<span>, while ultraviolets (UV) have frequency higher than these values (at the order of 1 PHz, </span>
).
The energy of electromagnetic radiation is proportional to its frequency, according to the equation
where h is the Planck constant and f is the frequency. We see that the higher the frequency, the greater the energy, so UV waves carry more energy than visible light.
Answer:
Equivalent
A conversion factor is a ratio of <u>e</u><u>quivalent</u> measures
Answer: E = 7,490.6 N/C
Explanation:
If we have a field E, and a particle with a charge q, the force that the particle experiences is:
F = E*q
In this case, we know that the force is:
F = 1.2*10^(-15) N
And we know that the particle is a proton, where the charge of a proton is:
q = 1.602*10^(-19) C
Then we can replace these two values in the equation to get:
1.2*10^(-15) N = E*1.602*10^(-19) C
We just need to isolate E.
(1.2*10^(-15) N)/(1.602*10^(-19) C) = E
7,490.6 N/C = E
That is the strength of the electric field.
Answer:
D. Friction
Explanation:
"Friction is caused by microwelds that form between two surfaces. Microwelds are stronger when the two surfaces are pushed together with a greater force." - ( Module 3 • Forces and Newton’s Laws) quoted
hope this helps and is right. p.s. i really need brainliest :)
Answer:
a) Magnitude = 1.03 m/s², Direction: south
b)
Explanation:
a) The magnitude and direction of the acceleration can be calculated using the following equation:
(1)
Where:
: is the final speed = 9.40 m/s
: is the initial speed = 13.0 m/s
t: is the time = 3.50 s
Solving equation (1) for a, we have:
Hence, the magnitude of the acceleration is 1.03 m/s² and the direction of the bird's acceleration is the opposite of the initial velocity direction, which means that the bird is decelerating.
b) The final velocity of the bird can be found using the same equation 1:
Therefore, the bird’s velocity after an additional 1.20 s has elapsed is 8.16 m/s.
I hope it helps you!