I think it occurs whenever waves come together so that they are in phase with each other.
C₂H₃O₂⁻ is an anion.
<u>Explanation:</u>
NaC₂H₃O₂(s) → Na⁺(aq) + C₂H₃O₂⁻(aq)
NaC₂H₃O₂ when dissociated, yields Na⁺ and C₂H₃O₂⁻.
Anion is a negatively charged ion.
In this case, C₂H₃O₂⁻ is an anion.
(a) +9.30 kg m/s
The impulse exerted on an object is equal to its change in momentum:

where
m is the mass of the object
is the change in velocity of the object, with
v = final velocity
u = initial velocity
For the volleyball in this problem:
m = 0.272 kg
u = -12.6 m/s
v = +21.6 m/s
So the impulse is

(b) 155 N
The impulse can also be rewritten as

where
F is the force exerted on the volleyball (which is equal and opposite to the force exerted by the volleyball on the fist of the player, according to Newton's third law)
is the duration of the collision
In this situation, we have

So we can re-arrange the equation to find the magnitude of the average force:

B. Transverse Wave this is the correct answer
Answer:
a
The orbital speed is 
b
The escape velocity of the rocket is 
Explanation:
Generally angular velocity is mathematically represented as
Where T is the period which is given as 1.6 days = 
Substituting the value


At the point when the rocket is on a circular orbit
The gravitational force = centripetal force and this can be mathematically represented as

Where G is the universal gravitational constant with a value 
M is the mass of the earth with a constant value of 
r is the distance between earth and circular orbit where the rocke is found
Making r the subject
![r = \sqrt[3]{\frac{GM}{w^2} }](https://tex.z-dn.net/?f=r%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7BGM%7D%7Bw%5E2%7D%20%7D)
![= \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }](https://tex.z-dn.net/?f=%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7B6.67%2A10%5E%7B-11%7D%20%2A%205.98%2A10%5E%7B24%7D%7D%7B%284.45%2A10%5E%7B-5%7D%29%5E2%7D%20%7D)

The orbital speed is represented mathematically as

Substituting value

The escape velocity is mathematically represented as

Substituting values

