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3 years ago

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A bell with a fundamental frequency of 880 Hz is moving toward an observer at 3.5 m/s. If the speed of sound is 343 m/s, what pi

tch would be heard by the observer?

1 answer:

Liula [17]3 years ago

3 0

Answer:

f = 889 Hz

Explanation:

As we know that when observer and source move relative to each other

Then in that case the frequency heard by the observer is different from actual frequency

The changed in the frequency is given as

$f = f_0(\frac{v}{v - v_s})$

now we know that

v = 343 m/s

$v_s = 3.5 m/s$

$f_0 = 880 Hz$

now we have

$f = 880(\frac{343}{343 - 3.5})$

$f = 889 Hz$

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A car traveling 34 mi/h accelerates uniformly for 4 s, covering 615 ft in this time. What was its acceleration? Round your answe

Contact [7]

Answer:

51.94 ft/s²

257.63 ft/s

Explanation:

t = Time taken = 4 s

u = Initial velocity = 34 mi/h

v = Final velocity

s = Displacement = 615 ft

a = Acceleration

Converting velocity to ft/s

$34\ mi/h=\frac{34\times 5280}{3600}=49.87\ ft/s$

Equation of motion

$s=ut+\frac{1}{2}at^2\\\Rightarrow a=2\frac{s-ut}{t^2}\\\Rightarrow a=2\left(\frac{615-49.87\times 4}{4^2}\right)\\\Rightarrow a=51.94\ ft/s^2$

Acceleration is 51.94 ft/s²

$v=u+at\\\Rightarrow v=49.87+51.94\times 4\\\Rightarrow v=257.63\ ft/s$

Final velocity at this time is 257.63 ft/s

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3 years ago

) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern that has the same frequency as the note tha

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(a) 0.0021 s, 2926.5 rad/s

The frequency of the B note is

$f= 466 Hz$

The time taken to make one complete cycle is equal to the period of the wave, which is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{466 Hz}=0.0021 s$

The angular frequency instead is given by

$\omega = 2\pi f$

And substituting

f = 466 Hz

We find

$\omega = 2\pi (466 Hz)=2926.5 rad/s$

(b) 20 Hz, 125.6 rad/s

In this case, the period of the sound wave is

T = 50.0 ms = 0.050 s

So the frequency is equal to the reciprocal of the period:

$f=\frac{1}{T}=\frac{1}{0.050 s}=20 Hz$

While the angular frequency is given by:

$\omega = 2\pi f = 2 \pi (20 Hz)=125.6 rad/s$

(c) $4.30\cdot 10^{14} Hz, 7.48\cdot 1^{14} Hz, 2.33\cdot 10^{-15} s, 1.34\cdot 10^{-15}s$

The minimum angular frequency of the light wave is

$\omega_1 = 2.7\cdot 10^{15}rad/s$

so the corresponding frequency is

$f=\frac{\omega}{2 \pi}=\frac{2.7\cdot 10^{15}rad/s}{2\pi}=4.30\cdot 10^{14} Hz$

and the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{4.30\cdot 10^{14}Hz}=2.33\cdot 10^{-15}s$

The maximum angular frequency of the light wave is

$\omega_2 = 4.7\cdot 10^{15}rad/s$

so the corresponding frequency is

$f=\frac{\omega}{2 \pi}=\frac{4.7\cdot 10^{15}rad/s}{2\pi}=7.48\cdot 10^{14} Hz$

and the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{7.48\cdot 10^{14}Hz}=1.34\cdot 10^{-15}s$

(d) $2.0\cdot 10^{-7}s, 3.14\cdot 10^{7} rad/s$

In this case, the frequency is

$f=5.0 MHz = 5.0 \cdot 10^6 Hz$

So the period in this case is

$T=\frac{1}{f}=\frac{1}{5.0\cdot 10^6 Hz}=2.0 \cdot 10^{-7} s$

While the angular frequency is given by

$\omega = 2\pi f=2 \pi (5.0\cdot 10^{6}Hz)=3.14\cdot 10^{7} rad/s$

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3 years ago

When a 5 kg rock is dropped from a height of 6 m on Planet X, it loses 24 J of GPE. What is the acceleration due to gravity on P

yan [13]

Answer:

g = 1.25m/s²

Explanation:

Given the following data;

Mass = 5kg

Height = 6m

Gravitational potential energy = 24J

To find the acceleration due to gravity;

Potential energy can be defined as an energy possessed by an object or body due to its position.

Mathematically, potential energy is given by the formula;

$P.E = mgh$

Where,

P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

GPE = mgh

Substituting into the equation, we have;

24 = 5*6*g

24 = 30g

g = 30/24

g = 1.25m/s²

Therefore, the acceleration due to gravity on Planet X is 1.25m/s².

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3 years ago

Which of the following is an example of how humans can increase biodiversity?

Leto [7]

Answer:

The following is an example of how humans can increase biodiversity

Provide Wildlife Corridors and Connections Between Green Spaces

Use Organic Maintenance Methods and Cut Back On Lawns

Use a Native Plant Palette and Plant Appropriately

Utilize Existing Green Space Connections

Be Mindful of Non-Native Predators

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2 years ago

A mass of 3.6 kg oscillate on a horizontal spring with a spring constant of 160 N/m.

Darya [45]

Answer:

48.7 J

Explanation:

For a mass-spring system, there is a continuous conversion of energy between elastic potential energy and kinetic energy.

In particular:

- The elastic potential energy is maximum when the system is at its maximum displacement

- The kinetic energy is maximum when the system passes through the equilibrium position

Therefore, the maximum kinetic energy of the system is given by:

$KE=\frac{1}{2}mv^2$

where

m is the mass

v is the speed at equilibrium position

In this problem:

m = 3.6 kg

v = 5.2 m/s

Therefore, the maximum kinetic energy is:

$KE=\frac{1}{2}(3.6)(5.2)^2=48.7 J$

6 0

3 years ago