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musickatia [10]
3 years ago
6

When the C4 key on a piano keyboard is pressed, a string inside the piano is struck by a hammer and begins vibrating back and fo

rth at approximately 260 cycles per second.
a. What is the frequency in Hertz of the sound wave?
b. What is the wavelength of the wave?
Physics
1 answer:
Anettt [7]3 years ago
8 0

Answer:

Frequency = 260 Hz and wavelength = 1.31 m

Explanation:

Given that,

A hammer begins vibrating back and forth at approximately 260 cycles per second.

(a) The frequency of an object is the number of vibrations per unit time. The frequency of the sound wave is 260 Hz.

(b) The speed of sound in air is 343 m/s. So,

v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{343}{260}\\\\\lambda=1.31\ m

Hence, this is the required solution.

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A 26.4 g silver ring (cp = 234 J/kg·°C) is heated to a temperature of 66.2°C and then placed in a calorimeter containing 4.94 ✕
Slav-nsk [51]

Answer:

The final temperature of the mixture = 64.834 °C.

Explanation:

Heat lost by the silver ring = heat gained by the water + heat transferred to the surrounding.

c₁m₁(t₁-t₃) = c₂m₂(t₃-t₂) + Q..............Equation 1

Where c₁ = specific heat capacity of the silver copper, m₁ = mass of the silver copper, t₁ = initial temperature of the silver copper, t₃ = final temperature of the mixture. c₂ = specific heat capacity of water, t₂ = initial temperature of water, m₂ = mass of water, Q = energy transferred to the surrounding.

making t₃ the subject of the equation,

t₃ = [c₁m₁t₁+c₂m₂t₂-Q]/(c₁m₁+c₂m₂)........................ Equation 2

Given: c₁ = 234 J/kg.°C, m₁ = 26.4 g, t₁ = 66.2 °C, c₂ = 4200 J/K.°C, m₂ = 4.92×10⁻² kg, t₂ = 24.0 °C, Q = 0.136 J.

Substituting into equation 2

t₃ = [(234×26.4×66.2)+(4200×0.0492×24)-0.136]/[(234×26.4)+(4200×0.0492)]

t₃ = (408957.12+4959.36-0.136)/(6177.6+206.64)

t₃ = (413916.48-0.136)/6384.24

t₃ = 413916.34/6384.24

Thus the final temperature of the mixture = 64.834 °C.

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3 years ago
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