(a) 0.0021 s, 2926.5 rad/s
The frequency of the B note is
The time taken to make one complete cycle is equal to the period of the wave, which is the reciprocal of the frequency:
The angular frequency instead is given by
And substituting
f = 466 Hz
We find
(b) 20 Hz, 125.6 rad/s
In this case, the period of the sound wave is
T = 50.0 ms = 0.050 s
So the frequency is equal to the reciprocal of the period:
While the angular frequency is given by:
(c) 
The minimum angular frequency of the light wave is
so the corresponding frequency is
and the period is the reciprocal of the frequency:
The maximum angular frequency of the light wave is
so the corresponding frequency is
and the period is the reciprocal of the frequency:
(d) 
In this case, the frequency is
So the period in this case is
While the angular frequency is given by
Answer:
0m/s²
Explanation:
Given parameters:
Initial velocity of the boat = 8m/s
Final velocity = 8m/s
Time taken = 4s
Unknown:
Acceleration of the boat = ?
Solution:
Acceleration is the rate of change of velocity with time.
It is mathematically expressed as;
A = 
A is the acceleration
v is the final velocity
u is the initial velocity
t is the time taken
Insert the parameters and solve;
A = 
= 0m/s²
I think the key here is to be exquisitely careful at all times, and
any time we make any move, keep our units with it.
We're given two angular speeds, and we need to solve for a time.
Outer (slower) planet:
Angular speed = ω rad/sec
Time per unit angle = (1/ω) sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/ω sec/rad) · (2π rad) = 2π/ω seconds .
Inner (faster) planet:
Angular speed = 2ω rad/sec
Time per unit angle = (1/2ω) sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/2ω sec/rad) · (2π rad) = 2π/2ω sec = π/ω seconds.
So far so good. We have the outer planet taking 2π/ω seconds for one
complete revolution, and the inner planet doing it in only π/ω seconds ...
half the time for double the angular speed. Perfect !
At this point, I know what I'm thinking, but it's hard to explain.
I'm pretty sure that the planets are in line on the same side whenever the
total elapsed time is something like a common multiple of their periods.
What I mean is:
They're in line, SOMEwhere on the circles, when
(a fraction of one orbit) = (the same fraction of the other orbit)
AND
the total elapsed time is a common multiple of their periods.
Wait ! Ignore all of that. I'm doing a good job of confusing myself, and
probably you too. It may be simpler than that. (I hope so.) Throw away
those last few paragraphs.
The planets are in line again as soon as the faster one has 'lapped'
the slower one ... gone around one more time.
So, however many of the longer period have passed, ONE MORE
of the shorter period have passed. We're just looking for the Least
Common Multiple of the two periods.
K (2π/ω seconds) = (K+1) (π/ω seconds)
2Kπ/ω = Kπ/ω + π/ω
Subtract Kπ/ω : Kπ/ω = π/ω
Multiply by ω/π : K = 1
(Now I have a feeling that I have just finished re-inventing the wheel.)
And there we have it:
In the time it takes the slower planet to revolve once,
the faster planet revolves twice, and catches up with it.
It will be 2π/ω seconds before the planets line up again.
When they do, they are again in the same position as shown
in the drawing.
To describe it another way . . .
When Kanye has completed its first revolution ...
Bieber has made it halfway around.
Bieber is crawling the rest of the way to the starting point while ...
Kanye is doing another complete revolution.
Kanye laps Bieber just as they both reach the starting point ...
Bieber for the first time, Kanye for the second time.
You're welcome. The generous bounty of 5 points is very gracious,
and is appreciated. The warm cloudy water and green breadcrust
are also delicious.
The force that prevents motion when the surfaces of two objects come into contact is known as friction. Friction decreases a machine's mechanical advantage, or, to put it another way, reduces the output to input ratio.
<h3>How can I figure out the frictional force?</h3>
The resistive force of friction (Fr) divided by the normal or perpendicular force (N) pushing the objects together yields the coefficient of friction (fr), which is a numerical value.
The formula fr = Fr/N serves as a representation of it.
Therefore, 100N of force is needed to move an item with a mass of 50 kg.
It will accelerate by 10 m/s2.
If a substance's mass does not change over time, friction cannot affect it. Instead, friction can be affected in a variety of ways by an object's mass.
To Learn more About Friction, Refer:
brainly.com/question/24338873
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