Answer:
F=50kg
D=5m
W=FD
therefore work done is 250Nm
Answer:
Momentum Packet Answer KEY - Science Online
YOU WILL SEE THE PDF
The distance between two consecutive nodes and the amplitude after 0.56s are m/2 and 1.75×10^(-4) m respectively.
<h3>What's the distance between consecutive nodes of the displacement of air molecules?</h3>
- Wavelength is the distance between two consecutive nodes or toughs or crests or anti-nodes.
- So, distance between consecutive nodes = wavelength = 2π÷k
= 2π/(4π÷m)
= m/2
<h3>What's the amplitude after 0.56s of the displacement of air molecules?</h3>
Displacement after 0.56 s = 0.008×cos(50π×0.56s)
=1.75×10^(-4) m
Thus, we can conclude that the distance between consecutive nodes and displacement after 0.56 s are m/2 and 1.75×10^(-4) m respectively.
Disclaimer: The question was given incomplete on the portal. Here is the complete question.
Question: The particle displacement y of air molecules due to a sound wave is given by y=0.008coswtsinkz where k=4π÷m and w=50π rads/s.
Calculate:
I) the distance between 2 consecutive nodes
ii) the amplitude after 0.565s
Learn more about the wavelength here:
brainly.com/question/10750459
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Answer:
M1 V1 = M1 V2 + M2 V3 conservation of momentum
V2 = (M1 V1 - M2 V3) / M1 where V2 = speed of M1 after impact
V2 = (3 * 9 - 1.5 * 5) / 9 = (27 - 7.5) / 9 = 2.17 m/s
Note: All speeds are in the same direction and have the same sign
Answer:
option (B) is the correct option.
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