Answer:
B. posititvely charged particles
Explanation:
Opposites attract to each other, and the same charge repels.
Answer:
The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.
Explanation:
Given that,
The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.
The speed of sound in air is 343 m/s.
To find,
The wavelength range for the corresponding frequency.
Solution,
The speed of sound is given by the following relation as :

Wavelength for f = 45 Hz is,


Wavelength for f = 375 Hz is,


So, the possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.
V = [4/3]π r^3 => [dV / dr ] = 4π r^2
[dV/dt] = [dV/dr] * [dr/dt]
[dV/dt] = [4π r^2] * [ dr/ dt]
r = 60 mm, [dr / dt] = 4 mm/s
[dV / dt ] = [4π(60mm)^2] * 4mm/s = 180,955.7 mm/s
C because the stack of paper was divided into 4