<span>The temporal lobe contains the area of the cortex involved in auditory processing called the primary auditory cortex.</span><span>
The temporal </span>lobe<span> is associated with </span>auditory processing<span> and olfaction.
</span>The primary auditory cortex<span> is the </span>part<span> of the temporal </span>lobe which<span> processes </span>auditory <span>information.</span>
Answer:
a) They are in the same point
b) t = 0 s, t = 2.27 s, t = 5.73 s
c) t = 1 s, t = 4.33 s
d) t = 2.67 s
Explanation:
Given equations are:
Constants are:
a) "Just after leaving the starting point" means that t = 0. So, if we look the equations, both 
and 
depend on t and don't have constant terms.
So both cars A and B are in the same point.
b) Firstly, they are in the same point in x = 0 at t = 0. But for generalized case, we must equalize equations and solve quadratic equation where roots will give us proper t value(s).
s, 
s
c) Since the distance isn't changing, the velocities are equal. To find velocities, we need to take the derivatives of both equations with respect to time and equalize them.
s, 
s
d) For same acceleration, we we need to take the derivatives of velocity equations with respect to time and equalize them.
s
Clever problem.
We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks. So if Fork-A is 256 Hz and the beat is 6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz. But which one is it ?
Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz. That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.
If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.
The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz. While it was loaded with wax, it was 261 Hz.
Answer: Helium has 2 protons.
