The task is to show that the right side of the equation has units of [Time], just like the left side has.
The right side of the equation is . . . 2 π √(L/G) .
We can completely ignore the 2π since it has no units at all, so it has no effect on the units of the right side of the equation. Now the task is simply to find the units of √(L/G) .
L . . . meters
G . . . meters/sec²
(L/G) = (meters) / (meters/sec²)
(L/G) = (meters) · (sec²/meters)
(L/G) = (meters · sec²) / (meters)
(L/G) = sec²
So √(L/G) = seconds = [Time]
THAT's what we were hoping to prove, and we did it !
Answer:
Well AM/Fm is digitally using soundwaves for cars through signal, which we cant hear without one (radio). Concerts are in real life and use wavelegnths that we hear because it bounces off overytihing.
Explanation:
Answer:
L = 130 decibels
Explanation:
The computation of the sound intensity level in decibels is shown below:
According to the question, data provided is as follows
I = sound intensity = 10 W/m^2
I0 = reference level = 
Now
Intensity level ( or Loudness)is
Therefore
L = 13 bel
And as we know that
1 bel = 10 decibels
So,
The Sound intensity level is
L = 130 decibels
Answer:
Vertical velocity decreases.
Explanation:
The motion of the ball is a projectile ball, which consists of two independent motions:
- a horizontal motion, with constant velocity
- a vertical motion, with constant acceleration g=9.8 m/s^2 towards the ground
In the vertical motion, there is a constant acceleration directed downward: this means that the vertical velocity decreases as the ball goes higher. In fact, it decreases following the equation
And it decreases until the ball reaches its maximum height, then it starts increasing again.
By using an electric field, it is feasible to differentiate between these different forms of radiation.
<h3>What is a radioactive source?</h3>
A source that emits radiation like gamma, beta, and alpha rays is said to be radioactive. Using an electric field, we can discriminate between these different forms of radiation.
The field does not deflate the gamma rays, but it does deflate the alpha and beta rays, with the alpha being deflated to the field's negative portion and the beta to its positive part.
Hence, by using an electric field, it is feasible to differentiate between these different forms of radiation.
To learn more about the radioactive source refer;
brainly.com/question/12741761
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