Answer:
a) The distance of spectator A to the player is 79.2 m
b) The distance of spectator B to the player is 43.9 m
c) The distance between the two spectators is 90.6 m
Explanation:
a) Knowing the time it takes the sound to reach both spectators, we can calculate their position relative to the player, using this equation:
x = v * t
where:
x = position of the spectators
v = speed of sound
t = time
Then, the position for spectator A relative to the player is:
x = 343 m/s * 0.231 s = 79.2 m
b)For spectator B:
x = 343 m/s * 0.128 s
x = 43.9 m
The distance of spectator A and B to the player is 79.2 m and 43.9 m respectively.
c) To calculate the distance between the spectators, please see the attached figure. Notice that the distance between the spectators is the hypotenuse of the triangle formed by the sightline of both. We already know the longitude of the two sides. Then, using Pythagoras theorem:
(Distance AB)² = A² + B²
(Distance AB)² = (79.2 m)² + (43.9 m)²
Distance AB = 90. 6 m
Because it is if you know you know and it is also helping the sentwcnde and air and confiscation
In rivers that are close to sea
Let
be the average acceleration over the first 2.46 seconds, and
the average acceleration over the next 6.79 seconds.
At the start, the car has velocity 30.0 m/s, and at the end of the total 9.25 second interval it has velocity 15.2 m/s. Let
be the velocity of the car after the first 2.46 seconds.
By definition of average acceleration, we have


and we're also told that

(or possibly the other way around; I'll consider that case later). We can solve for
in the ratio equation and substitute it into the first average acceleration equation, and in turn we end up with an equation independent of the accelerations:


Now we can solve for
. We find that

In the case that the ratio of accelerations is actually

we would instead have

in which case we would get a velocity of
