Answer:
When X = 0
Speed = maximum V (max) = ω A
Acceleration = zero a(max) = - ω^2 A
From x = A sin ω t sin = 0 so displacement = zero
V = ω A cos ω t cos = 1 and speed = maximum
a = - A ω^2 sin ω t sin = 0 and acceleration = zero
<span>(6.0x10^-22, -1.40x10^-21, 0) kg*m/s
Momentum is a conserved quantity. The total momentum of the system before and after the interactions will not change. So, let's look at the momentum before the interaction.
(3.2x10^-21, 0, 0) kg*m/s and (0,0,0) kg*m/s
After the interaction
(2.6x10^-21, 1.40x10^-21, 0) kg*m/s
and the other proton has to have a momentum that when added to this momentum equal the original value. Since the y and z vectors were initially 0, all we need for the y and x vector values of the result is to negate them. The x vector value will be
3.2x10^-21 - 2.6x10^-21 = 0.6x10^21 = 6.0x10^-22. So the other proton will have a momentum of
(6.0x10^-22, -1.40x10^-21, 0) kg*m/s</span>
Answer:
Option(b) is the correct answer to the given question .
Explanation:
The taste of the acids are sour due to H- ions where the taste of bases are bitter due to the OH-ions. The HP value of the acids are less than 7 where as the HP value of bases are greater than 7 .
The POH value of the acids are greater than the 7 and POH value of acids are less than the 7 . HP and the POH scale are used to find the difference between acid and the base.
- All the other options are not correct for the acids that's why these are incorrect option .
Therefore Option(b) is the correct answer .
