Answer:
750 J
Explanation:
lets convert mass into kg first , 150 /1000 = 0.15 kg
kinetic energy = =
4) Which of the following gases are typically used for colorful lighting when an electric current is applied ?
2 answers:
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Answer:
The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
Given that,
Amplitude = 0.08190 m
Frequency = 2.29 Hz
Wavelength = 1.87 m
(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave
Using formula of distance
Where, d = distance
A = amplitude
Put the value into the formula
Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.