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:
<em>The athlete will rise 1.10 meters off the ground</em>
Explanation:
<u>Vertical Motion</u>
If an object is launched vertically upwards at an initial speed vo, then it will reach a maximum height given by
The athlete can exert a net force upwards equal to twice his weight. It makes him accelerate upwards at
The speed at the end of his push can be computed by
Replacing the value of a obtained above:
where y is the length of this crouch
This is the initial speed of this vertical launch, thus
Answer:
(a) Steel rod:
Copper rod:
(b) Steel rod:
Copper rod:
Explanation:
Length of each rod = 0.75 m
Diameter of each rod = 1.50 cm = 0.015 m
Tensile force exerted = 4000 N
(a) Strain is given as the ratio of change in length to the original length of a body. Mathematically, it is given as
Strain =
where Y = Young modulus
F = Fore applied
A = Cross sectional area
For the steel rod:
Y = 200 000 000 000
F = 4000N
A = (r = d/2 = 0.015/2 = 0.0075 m)
=> A =
=> A = 0.000177
∴
For the copper rod:
Y = 120 000 000 000 N/m²
F = 4000N
A = (r = d/2 = 0.015/2 = 0.0075 m)
=> A =
=> A = 0.000177
(b) We can find the elongation by multiplying the Strain by the original length of the rods:
Elongation = Strain * Length
For the steel rod:
Elongation =
For the copper rod:
Elongation =
Answer:
The options are approximations of the exact answers:
A)
B)
C)
D) Toward the inner wall
E)
Explanation:
A) The electric field in a parallel plate capacitor is given by the formula , where
and in our case
and, for air,
, so we have:
B) The K+ ion has one elemental charge excess, so its charge is , and the force a charge experiments under an electric field E is given by F=qE, so we have:
C) The potential difference between two points separated a distance d under an uniform electric potential E is given by , so we have:
D) The electic field goes from positive to negative charges, so it goes towards the inner wall.
E) The work done by an electric field through a potential difference on a charge Q is
, and is equal to the kinetic energy imparted on it, so we have: