Velocity of an object of mass 5 kg increases from 3 m/s to 7 m/s on applying a force continuously for 2 s. Find out the force ap
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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