help mmmemememeeee I'M BEING TIMED

Complete Question:

A finch rides on the back of a Galapagos tortoise, which walks

at the stately pace of 0.060 m/s. After 1.5 minutes the finch tires of

the tortoise’s slow pace, and takes flight in the same direction for

another 1.5 minutes at 11 m/s.

What was the average speed of the  finch for this 3.0-minute interval?

Answer:

[tex]Speed = 5.53 m/s[/tex]

Explanation:

Distance is calculated as:

[tex]Distance = Speed * Time[/tex]

First, we calculate the distance for the first 1.5 minutes

For the first 1.5 minutes, we have:

[tex]Speed = 0.060m/s[/tex]

[tex]Time = 1.5\ mins[/tex]

[tex]D_2= 0.060m/s * 1.5\ mins[/tex]

Convert 1.5 mins to seconds

[tex]D_2= 0.060m/s * 1.5 * 60s[/tex]

[tex]D_2= 5.4m[/tex]

Next, we calculate the distance for the next 1.5 minutes

[tex]Speed = 11m/s[/tex]

[tex]Time = 1.5\ mins[/tex]

[tex]D_2= 11m/s * 1.5\ mins[/tex]

Convert 1.5 mins to seconds

[tex]D_2 = 11m/s * 1.5 * 60s[/tex]

[tex]D_2= 990m[/tex]

Total distance is:

[tex]Distance = 990m + 5.4m[/tex]

[tex]Distance = 995.4m[/tex]

The average speed for the 3.0 minute interval is:

[tex]Speed = \frac{Distance}{Time}[/tex]

[tex]Speed = \frac{995.4\ m}{3.0\ mins}[/tex]

Convert 3.0 minutes to seconds

[tex]Speed = \frac{995.4\ m}{3.0 * 60 secs}[/tex]

[tex]Speed = \frac{995.4\ m}{180 secs}[/tex]

[tex]Speed = 5.53 m/s[/tex]

Answer:

Explanation:

We shall apply Doppler's effect of sound .

speaker is the source , Jason is the observer . Source is moving at 10 m /s , observer is moving at  6 m/s .

apparent frequency = [tex]f_o\times\frac{V+v_o}{ V-v_s}[/tex]

V is velocity of sound , v₀ is velocity of observer and v_s is velocity of source and f_o is real frequency of source .

Here V = 340 m/s , v₀ is 6 m/s , v_s is 10 m/s . f_o = f

apparent frequency =  [tex]f\times \frac{340+6}{340-10}[/tex]

= [tex]f\times \frac{346}{330}[/tex]

So m = 346 , n = 330 .