Answer:
a) the distance between her and the wall is 13 m
b) the period of her up-and-down motion is 6.5 s
Explanation:
Given the data in the question;
wavelength λ = 26 m
velocity v = 4.0 m/s
a) How far from the wall is she?
Now, The first antinode is formed at a distance λ/2 from the wall, since the separation distance between the person and wall is;
x = λ/2
we substitute
x = 26 m / 2
x = 13 m
Therefore, the distance between her and the wall is 13 m
b) What is the period of her up-and-down motion?
we know that the relationship between frequency, wavelength and wave speed is;
v = fλ
hence, f = v/λ
we also know that frequency is expressed as the reciprocal of the time period;
f = 1/T
Hence
1/T = v/λ
solve for T
Tv = λ
T = λ/v
we substitute
T = 26 m / 4 m/s
T = 6.5 s
Therefore, the period of her up-and-down motion is 6.5 s
The average velocity of the car for the whole journey is 69.57 km/h.
The given parameters:
- <em>Length of the road, L = 320 km</em>
- <em>Distance covered = 240 km at 75 km/h</em>
- <em>time spent refueling, t₂ = 0.6 hr</em>
- <em>Final velocity, = 100 km/hr</em>
The time spent by the before refueling is calculated as follows;

The time spent by the car for the remaining journey;

The total time of the journey is calculated as follows;

The average velocity of the car for the whole journey is calculated as follows;

Learn more about average velocity here: brainly.com/question/6504879
Answer:
The speed the bat is gaining on its prey is 0.03m/s
Explanation:
Given;
speed of the bat, v₀ = 3.7 m/s
frequency of the bat, F₀ = 36 kHz
frequency of the source, Fs = 36.79
This is relative motion between a source of the sound and the observer. The phenomenon is known as Doppler effect.
Apply the following equation to determine the speed of the insect which is the source;
![F_0 = F_s[\frac{v+v_0}{v-v_s} ]\\\\\frac{F_0}{F_s} = [\frac{v+v_0}{v-v_s} ]\\\\\frac{36.79}{36} = \frac{340+3.7}{340-v_s}\\\\1.0219 = \frac{343.7}{340-v_s}\\\\ 340-v_s = \frac{343.7}{1.0219}\\\\340-v_s = 336.33\\\\v_s = 340-336.33\\\\v_s = 3.67 \ m/s](https://tex.z-dn.net/?f=F_0%20%3D%20F_s%5B%5Cfrac%7Bv%2Bv_0%7D%7Bv-v_s%7D%20%5D%5C%5C%5C%5C%5Cfrac%7BF_0%7D%7BF_s%7D%20%3D%20%5B%5Cfrac%7Bv%2Bv_0%7D%7Bv-v_s%7D%20%5D%5C%5C%5C%5C%5Cfrac%7B36.79%7D%7B36%7D%20%3D%20%5Cfrac%7B340%2B3.7%7D%7B340-v_s%7D%5C%5C%5C%5C1.0219%20%3D%20%5Cfrac%7B343.7%7D%7B340-v_s%7D%5C%5C%5C%5C%20%20340-v_s%20%3D%20%5Cfrac%7B343.7%7D%7B1.0219%7D%5C%5C%5C%5C340-v_s%20%3D%20336.33%5C%5C%5C%5Cv_s%20%3D%20340-336.33%5C%5C%5C%5Cv_s%20%3D%203.67%20%5C%20m%2Fs)
The speed the bat is gaining on its prey = 3.7m/s - 3.67m/s = 0.03 m/s
Therefore, the speed the bat is gaining on its prey is 0.03m/s