If the radius of the Earth decreases to one fourth (1/4) its original size, what would your NEW weight become
1. 4 times as much
2.1/4 as much
3.16 times as much
4.1/16 as much

Answer:524 Hz

Explanation:

Approximate frequency, heard in other car, when two car approaches each other, before they cross each other is 524 Hz.

Frequency of wave is the number of waves, which is passed thorough a particular point at a unit time.

For the two cars approaching each other the Doppler formula to find the frequency of second car is given as,

[tex]f_2=\dfrac{V_s+V_2}{V_s-V_1}f_1[/tex]

Here, [tex]V_s[/tex] is the speed of the sound.

Two cars approach each other from opposite directions.The velocity of car one is 54 km/h and the velocity of the car two is also 54 km/h.

Convert the unit of velocity of the car as,

[tex]\rm 54km/s=54\times\dfrac{5}{18}m/s\\\rm 54km/s=15m/s[/tex]

As we know that the speed of the sound is 340 m/s and one of the cars emits a note of frequency 480 Hz.

Thus, putting the values in the above formula to find out the frequency heard in the other car before they cross each other as,

[tex]f_2=\dfrac{340+15}{340-15}480\\f_2=524.3\rm Hz[/tex]

The approximate frequency heard in the other car before they cross each other is 524 Hz.

Learn more about the frequency here;

https://brainly.com/question/1436295

Answer:

v = 1.098 m/s

Explanation:

Given that,

A person walks 750m due north then 250 m due east of the entire walk takes 12 min.

We need to find the average velocity of the person.

Displacement,

[tex]d=\sqrt{750^2+250^2} \\\\d=790.56\ m[/tex]

Average velocity = displacement/time

[tex]v=\dfrac{790.56\ m}{12\times 60\ s}\\\\v=1.098\ m/s[/tex]

So, the average velocity of the person is 1.098 m/s.