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SpyIntel [72]
3 years ago
12

A whistle of frequency 564 Hz moves in a circle of radius 71.2 cm at an angular speed of 17.1 rad/s. What are (a) the lowest and

(b) the highest frequencies heard by a listener a long distance away, at rest with respect to the center of the circle? (Take the speed of sound in air to be 343 m/s.)
Physics
1 answer:
trapecia [35]3 years ago
3 0

Answer:

a) f'=544.66 \textup{Hz}

b) f'=584.75 \textup{Hz}

Explanation:

Given:

Frequency of the whistle, f = 564 Hz

Radius of the circle, r = 71.2 cm = 0.712 m

Angular speed, ω = 17.1 rad/s

speed of source, v_s = rω = 0.712 × 17.1 = 12.1752 m/s

speed of sound, v = 343 m/s

Now, applying the Doppler's effect formula, we have

f'=f\frac{v\pm v_d}{v\pm v_s}

where,

v_d = relative speed of the detector with respect to medium = 0

a) for lowest frequency, we have the formula as:

f'=f\frac{v}{v+v_s}

on substituting the values, we get

f'=564\times\frac{343}{343+12.1752}

or

f'=544.66 \textup{Hz}

b) for maximum frequency, we have the formula as:

f'=f\frac{v}{v-v_s}

on substituting the values, we get

f'=564\times\frac{343}{343-12.1752}

or

f'=584.75 \textup{Hz}

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Assume the acceleration of the object is a(t) = −9.8 meters per second per second. (Neglect air resistance.) With what initial v
lukranit [14]

Answer:

Vi = 94.64 m/s

Explanation:

I order to find out the initial velocity of the object, we can use third equation of motion:

2ah = Vf² - Vi²

where,

a = acceleration = -9.8 m/s²

h = maximum height covered by object = 460 m - 3 m = 457 m

Vf = Final Velocity = 0 m/s (since, object momentarily stops at highest point)

Vi = Initial Velocity = ?

Therefore,

2(-9.8 m/s²)(457 m) = (0 m/s)² - Vi²

Vi = √8957.2 m²/s²

<u>Vi = 94.64 m/s</u>

3 0
3 years ago
The radius of Earth is about 6450 km. A 7070 N spacecraft travels away from Earth. What is the weight of the spacecraft at a hei
Triss [41]

Answer:

(a) 1767.43 N

(b) 182.45 N

Explanation:

Radius of earth, R = 6450 km

Weight of person, W = 7070 N

mass of person, m = W / g = 7070 / 9.8 = 721.4 kg

(a) h = 6450 km

The value of acceleration due to gravity on height is given by

g' = g\left ( \frac{R}{R+h} \right )^2

g' = g\left ( \frac{6450}{6450+6450} \right )^2

g' = g / 4 = 9.8 / 4 = 2.45 m/s^2

The weight of the person at such height is

W' = m x g' = 721.4 x 2.45

W' = 1767.43 N

(b) h = 33700 km

The value of acceleration due to gravity on height is given by

g' = g\left ( \frac{R}{R+h} \right )^2

g' = g\left ( \frac{6450}{6450+33700} \right )^2

g' = g x 0.0258 = 9.8 x 0.0258 = 0.253 m/s^2

The weight of the person at such height is

W' = m x g'

W' = 721.4 x 0.253

W' = 182.45 N

3 0
3 years ago
Electrical wire with a diameter of .5 cm is wound on a spool with a radius of 30 cm and a height of 24 cm.
kow [346]

Answer:

a)   # lap = 301.59 rad , b)   L = 90.48 m

Explanation:

a) Let's use a direct proportions rule (rule of three). If one turn of the wire covers 0.05 cm, how many turns do you need to cover 24 cm

          # turns = 1 turn (24 cm / 0.5 cm)

         # laps = 48 laps

Let's reduce to radians

        # laps = 48 laps (2 round / 1 round)

       # lap = 301.59 rad

b) Each lap gives a length equal to the length of the circle

          L₀ = 2π R

          L = # turns L₀

          L = # turns 2π R

          L = 48 2π 30

          L = 9047.79 cm

          L = 90.48 m

6 0
3 years ago
Three 20.0 ohm resistors are
V125BC [204]

Answer:

6.67 ohm

Explanation:

From the question given above, the following data were obtained:

Resistor 1 (R₁) =20 ohm

Resistor 2 (R₂) = 20 ohm

Resistor 3 (R₃) = 20 ohm

Equivalent Resistance (R) =?

Since the resistors are arranged in parallel connection, the equivalent resistance can be obtained as follow:

1/R = 1/R₁ + 1/R₂ + 1/R₃

1/R = 1/20 + 1/20 + 1/20

1/R = (1 + 1 + 1) / 20

1/R = 3/20

Invert

R = 20/3

R = 6.67 ohm

Therefore, the equivalent resistance is 6.67 ohm.

5 0
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7 0
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