Answer:
68.8 Hz
137.6 Hz, 206.4 Hz
Explanation:
L = Length of tube = 2.5 m
v = Velocity of sound in air = 344 m/s
Distance between nodes is given by
Where n = 0, 1, 2, 3, ...
Making n+1 = n
where n = 1, 2, 3 .....
For fundamental frequency n = 1
Frequency is given by
The fundamental frequency is 68.8 Hz
First overtone
Second overtone
The overtones are 137.6 Hz, 206.4 Hz
Answer:
Explanation:
We shall solve this question with the help of Ampere's circuital law.
Ampere's ,law
∫ B dl = μ₀ I , B is magnetic field at distance x from the axis within wire
we shall find magnetic field at distance x . current enclosed in the area of circle of radius x
= I x π x² / π R²
= I x² / R²
B x 2π x = μ₀ x current enclosed
B x 2π x = μ₀ x I x² / R²
B = μ₀ I x / 2π R²
Maximum magnetic B₀ field will be when x = R
B₀ = μ₀I / 2π R
Given
B = B₀ / 3
μ₀ I x / 2π R² = μ₀I / 2π R x 3
x = R / 3
b ) The largest value of magnetic field is on the surface of wire
B₀ = μ₀I / 2π R
At distance x outside , let magnetic field be B
Applying Ampere's circuital law
∫ B dl = μ₀ I
B x 2π x = μ₀ I
B = μ₀ I / 2π x
Given B = B₀ / 3
μ₀ I / 2π x = μ₀I / 2π R x 3
x = 3R .
Answer:
B
Explanation:
The correct answer is B) have unlike charges. Since they are attracted to each other they have to be unlike
Answer:
Explanation:
Given data
Mass m=67.0 kg
Final Speed vf=8.00 m/s
Initial Speed vi=2.00 m/s
Distance d=25.0 m
Force F=30.0 N
From work-energy theorem we know that the work done equals the change in kinetic energy
W=ΔK=Kf-Ki=1/2mvf²-1/2mvi²
And
So
and we know that the force the sprinter exerted Fsprinter the force of the headwind Fwind=30.0N
So
