Fundamental frequency,
f=v2l=T/μ−−−−√2l
=(50)/0.1×10−3/10−22×0.6−−−−−−−−−−−−−−−−−−−√
=58.96Hz
Let, n th harmonic is the hightest frequency, then
(58.93)n = 20000
∴N=339.38
Hence, 339 is the highest frequency.
∴fmax=(339)(58.93)Hz=19977Hz.
<h3>
What is frequency?</h3>
In physics, frequency is the number of waves that pass a given point in a unit of time as well as the number of cycles or vibrations that a body in periodic motion experiences in a unit of time. After moving through a sequence of situations or locations and then returning to its initial position, a body in periodic motion is said to have experienced one cycle or one vibration. See also simple harmonic motion and angular velocity.
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All the boats operating at night requires Navigation light.
Navigation light helps prevent collisions between boats and see if visibility is poor. The types of boat are : Rowboats, Tug boats, Vessels, Sailboat etc.
Various boats have different lightning color to show its use and side of the boat.
Answer:
12.5752053801 m/s
No.
Explanation:
Q = Volume flow rate = 
d = Diameter of pipe = 9 cm
A = Area = 
Volume flow rate is given by
Velocity of fluid is 12.5752053801 m/s
The volume flow rate in m³/s is
The flow of fluid does not depend on the type of water used. Hence the answers would be same. If Q is constant v will be the same irrespective of the type of water used.
Answer:
The value of acceleration that accomplishes this is 8.61 ft/s² .
Explanation:
Given;
maximum distance to be traveled by the car when the brake is applied, d = 450 ft
initial velocity of the car, u = 60 mph = (1.467 x 60) = 88.02 ft/s
final velocity of the car when it stops, v = 0
Apply the following kinematic equation to solve for the deceleration of the car.
v² = u² + 2as
0 = 88.02² + (2 x 450)a
-900a = 7747.5204
a = -7747.5204 / 900
a = -8.61 ft/s²
|a| = 8.61 ft/s²
Therefore, the value of acceleration that accomplishes this is 8.61 ft/s² .
First,
where 
is density, 
is mass, and 
is volume. We can compute the volume of the roll:
When the roll is unfurled, the aluminum will be a rectangular box (a very thin one), so its volume will be the product of the given area and its thickness 
. Note that we're assuming the given area is not the actual total surface area of the aluminum box, but just the area of the largest face (i.e. the area of one side of the unrolled sheet of aluminum).
So we have
where 
is the given area, so
If we're taking significant digits into account, the volume we found would have been 
, in turn making the thickness 
.
