Sound travels at a speed of 330 meters/second. If Denise hears a police siren 150 meters away, approximately how long did it tak
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Answer:
Airplane speed relative to the ground is 260 km/h and θ = 22.6º direction from north to east
Explanation:
This is a problem of vector composition, a very practical method is to decompose the vectors with respect to an xy reference system, perform the sum of each component and then with the Pythagorean theorem and trigonometry find the result.
Let's take the north direction with the Y axis and the east direction as the X axis
Vy = 240 km / h airplane
Vx = 100 Km / h wind
a) See the annex
Analytical calculation of the magnitude of the speed and direction of the aircraft
V² = Vx² + Vy²
V = √ (240² + 100²)
V = 260 km/h
Airplane speed relative to the ground is 260 km/h
Tan θ = Vy / Vx
tan θ = 100/240
θ = 22.6º
Direction from north to eastb
b) What direction should the pilot have so that the resulting northbound
Vo = 240 km/h airplane
Vox = Vo cos θ
Voy = Vo sin θ
Vx = 100 km / h wind
To travel north the speeds the x axis (East) must add zero
Vx -Vox = 0
Vx = Vox = Vo cos θ
100 = 240 cos θ
θ = cos⁻¹ (100/240)
θ = 65.7º
North to West Direction
The speed in that case would be
V² = Vx² + Vy²
To go north we must find Vy
Vy² = V² - Vx²
Vy = √( 240² - 100²)
Vy = 218.2 km / h
(a) first two overtones for each string:
The first string has a fundamental frequency of 196 Hz. The n-th overtone corresponds to the (n+1)-th harmonic, which can be found by using
where f1 is the fundamental frequency.
So, the first overtone (2nd harmonic) of the string is
while the second overtone (3rd harmonic) is
Similarly, for the second string with fundamental frequency
, the first overtone is
and the second overtone is
(b) The fundamental frequency of a string is given by
where L is the string length, T the tension, and
is the mass per unit of length. This part of the problem says that the tension T and the length L of the string are the same, while the masses are different (let's calle them 
, the mass of the string of frequency 196 Hz, and 
, the mass of the string of frequency 523 Hz.
The ratio between the fundamental frequencies of the two strings is therefore:
and since L and T simplify in the equation, we can find the ratio between the two masses:
(c) Now the tension T and the mass per unit of length
is the same for the strings, while the lengths are different (let's call them 
and 
). Let's write again the ratio between the two fundamental frequencies
And since T and simplify, we get the ratio between the two lengths:
(d) Now the masses m and the lenghts L are the same, while the tensions are different (let's call them
and 
. Let's write again the ratio of the frequencies:
Now m and L simplify, and we get the ratio between the two tensions:
The first string has a fundamental frequency of 196 Hz. The n-th overtone corresponds to the (n+1)-th harmonic, which can be found by using
where f1 is the fundamental frequency.
So, the first overtone (2nd harmonic) of the string is
while the second overtone (3rd harmonic) is
Similarly, for the second string with fundamental frequency
and the second overtone is
(b) The fundamental frequency of a string is given by
where L is the string length, T the tension, and
The ratio between the fundamental frequencies of the two strings is therefore:
and since L and T simplify in the equation, we can find the ratio between the two masses:
(c) Now the tension T and the mass per unit of length
And since T and
(d) Now the masses m and the lenghts L are the same, while the tensions are different (let's call them
Now m and L simplify, and we get the ratio between the two tensions: