When air is blown into the open pipe,
L =
where nis any integral number 1,2,3,4 etc. and λ is the wavelength of the oscillation
⇒λ=
Note here that n=1 is for fundamental, n=2 is first harmonic and so on..
⇒ third harmonic will be n=4
Given L=6m, n=4, solving for λ we get:
λ= =3m
Relationship of frequency(f), velocity of sound (c) and wavelength(λ) is:
c=f.λ Or f=
⇒f=
≈115 Hz
Answer:
Therefore the required solution is
Explanation:
Given vibrating system is
Consider U(t) = A cosωt + B sinωt
Differentiating with respect to t
U'(t)= - A ω sinωt +B ω cos ωt
Again differentiating with respect to t
U''(t) = - A ω² cosωt -B ω² sin ωt
Putting this in given equation
Equating the coefficient of sinωt and cos ωt
.........(1)
and
........(2)
Solving equation (1) and (2) by cross multiplication method
and
Therefore the required solution is
Answer:
E) are almost circular, with low eccentricities.
Explanation:
Kepler's laws establish that:
All the planets revolve around the Sun in an elliptic orbit, with the Sun in one of the focus (Kepler's first law).
A planet describes equal areas in equal times (Kepler's second law).
The square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit (Kepler's third law).
Where T is the period of revolution and a is the semi-major axis.
Planets orbit around the Sun in an ellipse with the Sun in one of the focus. Because of that, it is not possible to the Sun to be at the center of the orbit, as the statement on option "C" says.
However, those orbits have low eccentricities (remember that an eccentricity = 0 corresponds to a circle)
In some moments of their orbit, planets will be closer to the Sun (known as perihelion). According with Kepler's second law to complete the same area in the same time, they have to speed up at their perihelion and slow down at their aphelion (point farther from the Sun in their orbit).
Therefore, option A and B can not be true.
In the celestial sphere, the path that the Sun moves in a period of a year is called ecliptic, and planets pass very closely to that path.
Answer:
46.4 s
Explanation:
5 minutes = 60 * 5 = 300 seconds
Let g = 9.8 m/s2. And be the slope of the road, s be the distance of the road, a be the acceleration generated by Rob, 3a/4 is the acceleration generated by Jim . Both of their motions are subjected to parallel component of the gravitational acceleration
Rob equation of motion can be modeled as s = a_Rt_R^2/2 = a300^2/2 = 45000a[/tex]
Jim equation of motion is
As both of them cover the same distance
So Jim should start 346.4 – 300 = 46.4 seconds earlier than Rob in other to reach the end at the same time