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3 years ago

At what step in the scientific method does a scientific propose the problem that he or she wants to solve

1 answer:

8 0

Answer: That would be the first step of the scientific method. After that, the scientist would research the problem and form a hypothesis

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A commuter train passes a passenger platform at a constant speed of 39.6 m/s. The train horn is sounded at its characteristic fr

Licemer1 [7]

Complete Question

A commuter train passes a passenger platform at a constant speed of 39.6 m/s. The train horn is sounded at its characteristic frequency of 350 Hz.

(a)

What overall change in frequency is detected by a person on the platform as the train moves from approaching to receding

(b) What wavelength is detected by a person on the platform as the train approaches?

Answer:

$\Delta f = 81.93 \ Hz$

$\lambda_1 = 0.867 \ m$

Explanation:

From the question we are told that

The speed of the train is $v_t = 39.6 m/s$

The frequency of the train horn is $f_t = 350 \ Hz$

Generally the speed of sound has a constant values of $v_s = 343 m/s$

Now according to dopplers equation when the train(source) approaches a person on the platform(observe) then the frequency on the sound observed by the observer can be mathematically represented as

$f_1 = f * \frac{v_s}{v_s - v_t}$

substituting values

$f_1 = 350 * \frac{343 }{343-39.6}$

$f_1 = 395.7 \ Hz$

Now according to dopplers equation when the train(source) moves away from the person on the platform(observe) then the frequency on the sound observed by the observer can be mathematically represented as

$f_2 = f * \frac{v_s}{v_s +v_t}$

substituting values

$f_2 = 350 * \frac{343}{343 + 39.6}$

$f_2 = 313.77 \ Hz$

The overall change in frequency is detected by a person on the platform as the train moves from approaching to receding is mathematically evaluated as

$\Delta f = f_1 - f_2$

$\Delta f = 395.7 - 313.77$

$\Delta f = 81.93 \ Hz$

Generally the wavelength detected by the person as the train approaches is mathematically represented as

$\lambda_1 = \frac{v}{f_1 }$

$\lambda_1 = \frac{343}{395.7 }$

$\lambda_1 = 0.867 \ m$

4 0

3 years ago

If the average pitcher is releasing the ball from a height of 1.8 m above the ground, and the pitcher's mound is 0.2 m higher th

neonofarm [45]

Answer: A. 2.0m above ground

Explanation:

Height of the pitcher from the floor of his mound = 1.8m

Height of the mound above the rest of the field = 0.2m

Since the height difference between the mound and the ground the catcher stands on = 0.2m and the pitcher throws the ball horizontally, then the catcher has to hold his glove at (1.8 + 0.2)m

above ground.

= 2.0m above ground (option A)

7 0

2 years ago

A 225-g object is attached to a spring that has a force constant of 74.5 N/m. The object is pulled 6.25 cm to the right of equil

snow_lady [41]

Answer:

1.137278672 m/s

+5.9 cm or -5.9 cm

Explanation:

A = Amplitude = 6.25 cm

m = Mass of object = 225 g

k = Spring constant = 74.5 N/m

Maximum speed is given by

$v_m=A\omega\\\Rightarrow v_m=A\sqrt{\dfrac{k}{m}}\\\Rightarrow v_m=6.25\times 10^{-2}\times \sqrt{\dfrac{74.5}{0.225}}\\\Rightarrow v_m=1.137278672\ m/s$

The maximum speed of the object is 1.137278672 m/s

Velocity is at any instant is given by

$\dfrac{v_m}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A\omega}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A}{3}=\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A^2}{9}=A^2-x^2\\\Rightarrow A^2-\dfrac{A^2}{9}=x^2\\\Rightarrow x^2=\dfrac{8}{9}A^2\\\Rightarrow x=A\sqrt{\dfrac{8}{9}}\\\Rightarrow x=6.25\times 10^{-2}\sqrt{\dfrac{8}{9}}\\\Rightarrow x=\pm 0.0589255650989\ m$

The locations are +5.9 cm or -5.9 cm

4 0

2 years ago

Velocity is intensive how?

White raven [17]

It is an intensive property as it varies with time and position within the system.

4 0

3 years ago

The longer the wavelength, the lower the frequency, and the lower the energy. Lesson 4.07 True False

jenyasd209 [6]

It’s true. longer wavelength means a lower frequency, and a shorter wavelength means a higher frequency!

3 0

2 years ago