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3 years ago

A car speeds past you sounding its horn at a frequency of 1000Hz. The frequency you hear:

Physics

1 answer:

ki77a [65]3 years ago

3 0

Answer:

f' =1064.1025 Hz

Explanation:

We know that from doppler's effect

$f'=\frac{v+v_0}{v-v_s}f$

f'= observed frequency

f= actual frequency =1000 Hz

v= velocity of sound waves = 332 m/s

v_o= velocity of observer =0

v_s= velocity of source. ( let velocity of car be 20 m/s)

therefore,

$f'=\frac{332+0}{332-20}\times1000$

f' =1064.1025 Hz

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A tuning fork generates sound waves with a frequency of 240 Hz. The waves travel in opposite directions along a hallway, are ref

Bingel [31]

Answer:

The phase difference between the reflected waves when they meet at the tuning fork is 159.29 rad.

Explanation:

Given that,

Frequency of sound wave = 240 Hz

Distance = 46.0 m

Distance of fork = 14 .0 m

We need to calculate the path difference

Using formula of path difference

$\Delta x=2(L_{2}-L_{1})$

Put the value into the formula

$\Delta x =2((46.0-14.0)-14.0)$

$\Delta x=36\ m$

We need to calculate the wavelength

Using formula of wavelength

$\lambda=\dfrac{v}{f}$

Put the value into the formula

$\lambda=\dfrac{343}{240}$

$\lambda=1.42\ m$

We need to calculate the phase difference

Using formula of the phase difference

$\phi=\dfrac{2\pi}{\lambda}\times \delta x$

Put the value into the formula

$\phi=\dfrac{2\pi}{1.42}\times36$

$\phi=159.29\ rad$

$\phi\approx 68.2^{\circ}$

Hence, The phase difference between the reflected waves when they meet at the tuning fork is 159.29 rad.

7 0

3 years ago

Two children are pulling on opposite sides of a blanket. The brother is pulling with a force of 3 N. The sister is pulling with

Citrus2011 [14]

Let F1=Force exerted by the brother (+F1)
F1= Force exerted by the sister (-F2)

Fnet=(+F1) + (-F2)
Fnet= (+F1) + (-F2)
Fnet=F1 - F2
Fnet= (+3N)+(-5N)
Fnet= -2N

-F

towards the sister (-F) (greater force applied)

7 0

3 years ago

What is the half-life for the radioactive decay of particle that

pogonyaev

Answer:

ytyug

Explanation:

4 0

3 years ago

A 0.43 kg hammer is moving horizontally at 5.2 m/s when it strikes a nail and comes to rest after driving the nail 0.014 m into

vagabundo [1.1K]

Answer:

The duration of the impact is 0.005384 seconds

Explanation:

Given

m = 0.43 kg

v = 5.2 m/s

x = 0.014 m

Knowing the formulas

$v^{2}_{f} = v^{2}_{i} + 2ax\\0 = 5.2^{2} + 2a*0.014\\a = - 965.71 m/s^{2} \\\\vf = vi + at\\0 = 5.2 + (965.71)t\\t = 0.005384 s$

5 0

3 years ago

Read 2 more answers

Two teams of nine members each engage in tug-of-war. Each of the first team's members has an average mass of 68 kg and exerts an

diamong [38]

Answer:

(a) Acceleration = 0.1063 m/s^2 (Second team wins)

(b) Tension in rope = 65.106 N

Explanation:

Total mass of first team = 68 * 9 = 612 kg

Total force of first team = 1350 * 9 = 12150 N

Total mass of second team = 73 * 9 = 657 kg

Total force of seconds team = 1365 * 9 = 12285 N

Difference in force = 12285 - 12150 = 135 N (towards the second team as it has more force)

(a) For acceleration we get:

F = m * a

135 = (mass of both teams) * a

a = 135 / (612 + 657)

acceleration = 0.1063 m/s^2 (Second team wins)

(b) Since we know the acceleration of the first team (pulling being pulled towards the second team at an acceleration of 0.1063 m/s^2) , we can find out the force required to move them:

Force required for first team = mass of first team * acceleration

Force required = 612 * 0.1063

Force required = 65.106 N

This is the force exerted on the first team through the rope, so the tension in the rope will also be 65.106 N.

7 0

4 years ago