The answer is B. I don’t think I need to explain this,
Mean is average, Mode is the most common number, and Median is the middle number when you put the numbers is numerical order from least to greatest
Answer:
The ball will have an upward velocity of 6 m/s at a height of 5.51 m.
Explanation:
Hi there!
The equations of height and velocity of the ball are the following:
y = y0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
y = height at time t.
y0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).
v = velocity of the ball at time t.
Placing the origin at the throwing point, y0 = 0.
Let´s use the equation of velocity to obtain the time at which the velocity is 12.0 m/s / 2 = 6.00 m/s.
v = v0 + g · t
6.00 m/s = 12.0 m/s -9.81 m/s² · t
(6.00 - 12.0)m/s / -9.81 m/s² = t
t = 0.612 s
Now, let´s calculate the height of the baseball at that time:
y = y0 + v0 · t + 1/2 · g · t² (y0 = 0)
y = 12.0 m/s · 0.612 s - 1/2 · 9.81 m/s² · (0.612 s)²
y = 5.51 m
The ball will have an upward velocity of 6 m/s at a height of 5.51 m.
Have a nice day!
As the question is about changing in frequency of a wave for an observer who is moving relative to the wave source, the concept that should come to our minds is "
Doppler's effect."
Now the general formula of the Doppler's effect is:

-- (A)
Note: We do not need to worry about the signs, as everything is moving towards each other. If something/somebody were moving away, we would have the negative sign. However, in this problem it is not the issue.
Where,
g = Speed of sound = 340m/s.

= Velocity of the receiver/observer relative to the medium = ?.

= Velocity of the source with respect to medium = 0 m/s.

= Frequency emitted from source = 400 Hz.

= Observed frequency = 408Hz.
Plug-in the above values in the equation (A), you would get:


Solving above would give you,

= 6.8 m/s
The correct answer = 6.8m/s
We solve this using special
relativity. Special relativity actually places the relativistic mass to be the
rest mass factored by a constant "gamma". The gamma is equal to 1/sqrt
(1 - (v/c)^2). <span>
We want a ratio of 3000000 to 1, or 3 million to 1.
</span>
<span>Therefore:
3E6 = 1/sqrt (1 - (v/c)^2)
1 - (v/c)^2 = (0.000000333)^2
0.99999999999999 = (v/c)^2
0.99999999999999 = v/c
<span>v= 99.999999999999% of the speed of light ~ speed of light
<span>v = 3 x 10^8 m/s</span></span></span>