A 4-m long wire with a mass of 70 g is under tension. A transverse wave for which the frequency is 300 Hz, the wavelength is 0.6
m, and the amplitude is 6 mm is propagating on the wire. The maximum transverse acceleration of a point on a wire is closest to
1 answer:
Answer:5400 m/s^2
Explanation: In order to calculate the acceleration we have to use the centripetal form of it given by:
a= v^2/R where v is the velocity and r the radius of the circular movement which correspond to the maximum of amplitude of the transverse wave.
To calculate the velocity we know that v=ν*λ, where νand λ, are the frequency and wavelength of the wave, respectively. v=300*0.6=180 m/s
Finally we have:
a= (180 m/s)^2/6 m= 5400 m/s^2
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Solution:
f ( t )= 20 S ( t ) + 55/30 tS ( t )− 55/30 ( t − 30 ) S ( t − 30 )
• Taking the Laplace Transform:
F ( s ) = 20/s + 55/30 ( 1/s^2 ) – 55/30 ( 1/s^2) e^-30s = 20/s + 55/30 ( 1/s^2 ) ( 1 – e^-30s)
Answer:
<u>Resolving</u><u> </u><u>horizontally</u><u>.</u> :

therefore, for resultant:

substitute:

I have the exact same question, any chance you figured it out since you posted this?
Formula for potassium dichromate is
K2Cr2O7