Answer:
The length of rod A will be <u>greater than </u>the length of rod B
Explanation:
We, know that the formula for final length in linear thermal expansion of a rod is:
L' = L(1 + ∝ΔT)
where,
L' = Final Length
L = Initial Length
∝ = Co-efficient of linear expansion
ΔT = Change in temperature
Since, the rods here have same original length and the temperature difference is same as well. Therefore, the final length will only depend upon the coefficient of linear expansion.
For Rod A:
∝₁ = 12 x 10⁻⁶ °C⁻¹
For Rod B:
∝₂ = β₂/3
where,
β₂ = Coefficient of volumetric expansion for rod B = 24 x 10⁻⁶ °C⁻¹
Therefore,
∝₂ = 24 x 10⁻⁶ °C⁻¹/3
∝₂ = 8 x 10⁻⁶ °C⁻¹
Since,
∝₁ > ∝₂
Therefore,
L₁ > L₂
So, the length of rod A will be <u>greater than </u>the length of rod B
Answer: A = square root (2/L)
Explanation: find the attached file for explanation
Answer:
a.241.08 m/s b. 196 Hz c. 392 Hz
Explanation:
a. Determine the speed of waves within the wire.
The frequency of oscillation of the wave in the string, f = nv/2L where n = harmonic number, v = speed of wave in string, L = length of string = 1.23 m.
Since f = 588 Hz which is the 6 th harmonic, n = 6. So, making v subject of the formula, we have
v = 2Lf/n
substituting the values of the variables into v. we have
v = 2 × 1.23 m × 588Hz/6
v = 241.08 m/s
b. Determine the frequency at which the wire will vibrate with the first harmonic wave pattern.
The first harmonic is obtained from f when n = 1,
So, f = v/2L = 241.08 m/s ÷ 1.23m = 196 Hz
c. Determine the frequency at which the wire will vibrate with the second harmonic wave pattern.
The second harmonic f' = 2f = 2 × 196 Hz = 392 Hz
Answer:
Displacement
Explanation:
The area under a velocity-time graph is the displacement. Velocity can be negative if an object is moving backwards.
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