Identical particles are placed at the 50-cm and 80-cm marks on a meter stick of negligible mass. This rigid body is then mounted
1 answer:
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Answer:
1. λ = 2 L, 2. v = 2L f₁ , 3. v = √ T /μ², 4. μ = 2,287 10⁻³ kg / m , 5. Δv / v = 0.058 , 6. Δμ / μ = 0.12 , 7. Δ μ = 0.3 10⁻³ kg / m ,
8. μ = (2.3 ±0.3) 10⁻³ kg / m
Explanation:
The speed of a wave is
v = λ f 1
Where f is the frequency and λ the wavelength
The speed is given by the physical quantities of the system with the expression
v = √ T /μ² 2
1) The fundamental frequency of a string is when at the ends we have nodes and a maximum in the center, therefore this is
L = λ / 2
λ = 2 L
2) For this we substitute in equation 1
v = 2L f₁
3) let's clear from equation 2
The speed of a wave is
v = λ f₁
Where f is the frequency and Lam the wavelength
The speed is given by the physical quantities of the system with the expression
v = √ T /μ² 2
4) linear density is
μ = T / (2 L f₁)²
μ = 5.08 / (2 0.812 29.02)²
μ = 2,287 10⁻³ kg / m
We maintain three significant length figures, so the result is reduced to
μ = 2.29 10⁻³ kg / m
5) the speed of the wave is
v = 2 L f₁
The fractional uncertainty is
Δv / v = ΔL / L + Δf₁ / F₁
Δv / v = 0.02 / 0.812 + 1 / 29.02
Δv / v = 0.024 + 0.034
Δv / v = 0.058
6) the equation for linear density is
μ = T / (2 L f₁)²
Δμ / μ = 2 ΔL / L + 2Δf₁ / f₁
The tension is an exact value therefore its uncertainty is zero ΔT = 0
Δμ / μ = 2 0.02 / 0.812 + 2 1 / 29.02
Δμ / μ = 0.12
7) absolute uncertainty
Δ μ = μ
Δ μ = 0.12 2.29 10⁻³ kg / m
Δ μ = 0.3 10⁻³ kg / m
8)
μ = (2.3 ±0.3) 10⁻³ kg / m
Answer:
Vf = 210 [m/s]
Av = 105 [m/s]
y = 2205 [m]
Explanation:
To solve this problem we must use the following formula of kinematics.
where:
Vf = final velocity [m/s]
Vo = initial velocity = 0 (released from the rest)
g = gravity acceleration = 10 [m/s²]
t = time = 21 [s]
Vf = 0 + (10*21)
Vf = 210 [m/s]
Note: The positive sign for the gravity acceleration means that the object is falling in the same direction of the gravity acceleration (downwards)
The average speed is defined as the sum of the final speed plus the initial speed divided by two. (the initial velocity is zero)
Av = (210 + 0)/2
Av = 105 [m/s]
To calculate the distance we must use the following equation of kinematics
44100 = 20*y
y = 2205 [m]