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Hoochie [10]
3 years ago
6

A student using a tuning fork of frequency 512 Hz observes that the speed of sound is 340.0 m/s. What is the wavelength of this

sound wave
Physics
1 answer:
Romashka [77]3 years ago
5 0

Answer:

0.066m

Explanation:

Step one:

given

frequency =512 Hz

The speed of sound is 340.0 m/s.

Required

The wavelength

Step two:

the formula for wavelength is

v=f \lambda

\lambda= v/f

substitute the given data

\lambda= 340/514\\\\\lambda= 0.66m

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Two speakers separated by a distance of 4.40 m emit sound. The speakers have opposite phase. A person listens from a location 3.
Hoochie [10]

Answer:

f = 147.21 Hz

Explanation:

In order to have a destructive interference, as the source emit in opposite phases, the path difference between the distance to the person, measured in a straight line from the speakers, must be equal to an integer number of wavelengths.

We need to know the distance from the listener to the other speaker, located 4.4 m from the one which is directly in front of him, which we can find using Pythagorean theorem, as follows:

l₂ = √(3)²+(4.4)² = 5.33 m

The difference in path will be, then:

d = l₂-l₁ = 5.33 m - 3.00 m = 2.33 m

For the lowest frequency that gives destructive interference, the wavelength will be highest possible, which happens when the distance is just one wavelength.

⇒ d = λ = 2.33 m

In any wave, there exists a fixed relationship between speed, frequency and wavelength, as follows:

v = λ*f Κ  ⇒ f = v/λ

Taking the speed of sound as 343 m/s, and solving for f, we get:

f= 343 m/s / 2.33 m = 147.21 Hz

3 0
3 years ago
The cheetah is considered the fastest running animal in the world. Cheetahs can accelerate to a speed of 21.7 m/s in 2.50 s and
viktelen [127]

Answer:

1) 64.2 mi/h

2) 3.31 seconds

3) 47.5 m

4) 5.26 seconds

Explanation:

t = Time taken = 2.5 s

u = Initial velocity = 0 m/s

v = Final velocity = 21.7 m/s

s = Displacement

a = Acceleration

1) Top speed = 28.7 m/s

1 mile = 1609.344 m

1\ m=\frac{1}{1609.344}\ miles

1 hour = 60×60 seconds

1\ s=\frac{1}{3600}\ hours

28.7\ m/s=\frac{\frac{28.7}{1609.344}}{\frac{1}{3600}}=64.2\ mi/h

Top speed of the cheetah is 64.2 mi/h

Equation of motion

v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow t=\frac{21.7-0}{2.5}\\\Rightarrow a=8.68\ m/s^2

Acceleration of the cheetah is 8.68 m/s²

2)

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{28.7-0}{8.68}\\\Rightarrow t=3.31\ s

It takes a cheetah 3.31 seconds to reach its top speed.

3)

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{28.7^2-0^2}{2\times 8.68}\\\Rightarrow s=47.5\ m

It travels 47.5 m in that time

4) When s = 120 m

s=ut+\frac{1}{2}at^2\\\Rightarrow 120=0\times t+\frac{1}{2}\times 8.68\times t^2\\\Rightarrow t=\sqrt{\frac{120\times 2}{8.68}}\\\Rightarrow t=5.26\ s

The time it takes the cheetah to reach a rabbit is 120 m is 5.26 seconds

8 0
3 years ago
Which type of wave is faster?/Seismic Waves
Diano4ka-milaya [45]

Answer:

me

Explanation:

7 0
3 years ago
Read 2 more answers
Help please it's urgent
insens350 [35]
I cant read it, i could most likely help if i could read it.
7 0
3 years ago
What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of – 30.0 nC?
Nastasia [14]

Answer:

F=1.26*10^{-3}N

Explanation:

Assuming the pith balls as point charges, we can calculate the repulsive force between them, using Coulomb's law:

F=\frac{kq_1q_2}{d^2}

We observe that the magnitude of the electric force is directly proportional to the product of the magnitude of both signed charges(q_1,q_2) and inversely proportional to the square of the distance(d) that separates them.

Replacing the given values, where k is the Coulomb constant:

F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-30*10^{-9}C)(-30*10^{-9}C)}{(8*10^{-2}m)^2}\\F=1.26*10^{-3}N

8 0
3 years ago
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