Answer:
Densities increase down the group
MP and BP decrease down the group
Softness increased going down the group
Speed of reacting increases going down the group
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.
The answer is b.) the momentum before the collision is greater than the momentum after the collision
For the answer to the question above, we'll have to use these formulas.
A) to find time to travel the 300m,
just find horizontal component of the velocity and divide.
ie x=89 x t x cos 40, t=x/89 x cos 40
<span>B) y=vtsin 40 - gt^2/2, just sub in
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I believe you can do the rest.
I hope I helped you with my answers.