A stretched string is 120 cm long and has a linear density of 0.022 g/cm. What tension in the string will result in a second har
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Answer
given,
Length of the string, L = 2 m
speed of the wave , v = 50 m/s
string is stretched between two string
For the waves the nodes must be between the strings
the wavelength is given by
where n is the number of antinodes; n = 1,2,3,...
the frequency expression is given by
now, wavelength calculation
n = 1
λ₁ = 4 m
n = 2
λ₂ = 2 m
n =3
λ₃ = 1.333 m
now, frequency calculation
n = 1
f₁ = 12.5 Hz
n = 2
f₂= 25 Hz
n = 3
f₃ = 37.5 Hz
Initially, the spring stretches by 3 cm under a force of 15 N. From these data, we can find the value of the spring constant, given by Hook's law:
where F is the force applied, and
is the stretch of the spring with respect to its equilibrium position. Using the data, we find
Now a force of 30 N is applied to the same spring, with constant k=5.0 N/cm. Using again Hook's law, we can find the new stretch of the spring:
where F is the force applied, and
Now a force of 30 N is applied to the same spring, with constant k=5.0 N/cm. Using again Hook's law, we can find the new stretch of the spring:
Answer:
L = μ₀ n r / 2I
Explanation:
This exercise we must relate several equations, let's start writing the voltage in a coil
= - L dI / dt
Let's use Faraday's law
E = - d Ф_B / dt
in the case of the coil this voltage is the same, so we can equal the two relationships
- d Ф_B / dt = - L dI / dt
The magnetic flux is the sum of the flux in each turn, if there are n turns in the coil
n d Ф_B = L dI
we can remove the differentials
n Ф_B = L I
magnetic flux is defined by
Ф_B = B . A
in this case the direction of the magnetic field is along the coil and the normal direction to the area as well, therefore the scalar product is reduced to the algebraic product
n B A = L I
the loop area is
A = π R²
we substitute
n B π R² = L I (1)
To find the magnetic field in the coil let's use Ampere's law
∫ B. ds = μ₀ I
where B is the magnetic field and s is the current circulation, in the coil the current circulates along the length of the coil
s = 2π R
we solve
B 2ππ R = μ₀ I
B = μ₀ I / 2πR
we substitute in
n ( μ₀ I / 2πR) π R² = L I
n μ₀ R / 2 = L I
L = μ₀ n r / 2I