If you are choosing to get a digital photo album it can come with the advantages like, it’s more economically friendly because you’re not having to print out pictures. But it’s disadvantage is that you can lose it easily or accidentally delete it some way. You also have a choice of a physical photo album, it’s advantages are that you won’t lose it as easily and they might last longer than a digital one’s. It’s disadvantage though is that with all the pictures and the book to keep the pictures in takes a lot from the economy. A solution to the digital photo album is to save in on a hard drive or something like that so you can keep it without having to worry about it being deleted as easily. The solution with the physical photo album is that you could get a photo holder that you can slide picture into pockets because it will keep the pictures from damage as well as using less paper products.
you can read over it and change things i might have done wrong and if i completely missed the point of this i’m sorry and hope you have a good day/night. (also it’s 6 sentences)
Mechanical energy
I think
Answer:
d₂ = 1.466 m
Explanation:
In this case we must use the rotational equilibrium equations
Στ = 0
τ = F r
we must set a reference system, we use with origin at the easel B and an axis parallel to the plank
, we will use that the counterclockwise ratio is positive
+ W d₁ - w_cat d₂ = 0
d₂ = W / w d₁
d₂ = M /m d₁
d₂ = 5.00 /2.9 0.850
d₂ = 1.466 m
Kepler's second law of planetary motion says that the imaginary line
from the Sun to the planet sweeps out equal areas in equal periods
of time.
It's just as true of comets, asteroids, dwarfs, coasting space probes ...
anything in the solar system that's coasting in gravity, and not firing engines.
That's why a comet moves so fast when it gets near the Sun.
This is a Doppler effect. Generally, if you move to a frequency source, you would detect an increase in frequency and when you move away from a source you would detect a decrease.
For this question, before you pass them, you are actually approaching them, so you would hear a higher frequency than the constant 300 Hz they are playing at.
Using the condensed formula:
f ' = ((v <u>+</u> vd)/(v <u>+</u> vs)) * f
Where: vd = Velocity of the detector.
vs = Velocity of the frequency source.
v = Velocity of sound in air.
f ' = Apparent frequency.
f = Frequency of source.
v = 343 m/s, vd = detector = 27.8 m/s, vs = velocity of the source =0. (the flautists are not moving).
f = 300 Hz.
There would be an overall increase in frequency, so we maintain a plus at the numerator and a minus at the denominator.
f ' = ((v + vd)/(v - vs)) * f
f ' = ((343+ 27.8)/(343 - 0)) * 300
= (370.8/343)* 300 = 324.3
Therefore frequency before passing them = 324.3 Hz.
Cheers.
