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.
These are the characteristics that apply:
- In a solution taste sour: which is consequence of the H+ concentration.
- Corrode metals: the H+ ion reacts with the metal producing a salt and water
-Produce hydronium ion in solution: as per the Bronsted - Lowry definition an acid is a substance that donates a proton, H+. This proton will react with H2O to form H3O+ (hydronium), as per this scheme:
HA + H2O --> A(-) + H3O(+)
The answer is in the attachment
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Answer:
16km
Explanation:
First change the minutes into hours then multiply by the distance.
(8÷60)×120=16km
If the pulling is done parallel to the floor with constant velocity, then the box is in equilibrium. In particular, the weight and normal force cancel, so that
<em>n</em> = 38 N
The friction force is proportional to the normal force by a factor of 0.27, so that
<em>f</em> = 0.27 (38 N) ≈ 10.3 N
and so the answer is D.
