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2 years ago

una onda longitudinal tiene una frecuencia de 200 hz y una longitud de onda de 4.2m ¿cual es la rapidez de la onda?

Physics

1 answer:

swat322 years ago

4 0

Answer:

v = 8.4 m/s

Explanation:

The question ays, "A longitudinal wave has a frequency of 200 Hz and a wavelength of 4.2m. What is the speed of the wave?".

Frequency of a wave, f = 200 Hz

Wavelength = 4.2 cm = 0.042 m

We need to find the speed of the wave. The formula for the speed of a wave is given by :

$v=f\lambda\\\\v=200\times 0.042\\\\=8.4\ m/s$

So, the speed of the wave is equal to 8.4 m/s.

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Ocean waves pass through two small openings, 20.0 m apart, in a breakwater. You're in a boat 70.0 m from the breakwater and init

Klio2033 [76]

Answer:

λ = 5.65m

Explanation:

The Path Difference Condition is given as:

δ= $(m+\frac{1}{2})\frac{lamda}{n}$ ;

where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.

m = no of openings which is 2

∴δ= $\frac{3*lamda}{2}$

n is the index of refraction of the medium in which the wave is traveling

To find δ we have;

δ= $\sqrt{70^2+(33+\frac{20}{2})^2 }-\sqrt{70^2+(33-\frac{20}{2})^2 }$

δ= $\sqrt{4900+(\frac{66+20}{2})^2}-\sqrt{4900+(\frac{66-20}{2})^2}$

δ= $\sqrt{4900+(\frac{86}{2})^2 }-\sqrt{4900+(\frac{46}{2})^2 }$

δ= $\sqrt{4900+43^2}-\sqrt{4900+23^2}$

δ= $\sqrt{4900+1849}-\sqrt{4900+529}$

δ= $\sqrt{6749}-\sqrt{5429}$

δ= 82.15 -73.68

δ= 8.47

Again remember; to calculate the wavelength of the ocean waves; we have:

δ= $\frac{3*lamda}{2}$

δ= 8.47

8.47 = $\frac{3*lamda}{2}$

λ = $\frac{8.47*2}{3}$

λ = 5.65m

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3 years ago

When jogging outside you accidently bump into a curb. Your feet stop but your body continues to move forward and you end up on t

oee [108]

Inertia I think because I've heard it around school and in science

6 0

2 years ago

Why do we need to learn about median and mode?

siniylev [52]

We need to learn about median and mode because they are both important in statistical data and statistical data is what shows us important information about the world around us.

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3 years ago

A rod extending between x = 0 and x = 13.0 cm has uniform cross-sectional area A = 8.00 cm2. Its density increases steadily betw

mezya [45]

Answer:

The mass is $m = 3.45408 kg$

Explanation:

From the question we are told that

The extension of the rod is $from , \ x_1 = 0 \to x_2 = 13.0$

The area is $A = 8.0 cm^2$

The density increase as follows $from \ \rho_1 =2.5 g/cm^2 \to \rho_2= 19.0 g/cm^3$

The equation $\rho = B + Cx$

at $x_1= 0$ $\rho_1 =2.5 g/cm^2$

$2.5 = B + 0$

=> $B =2.5$

So at $x_2 = 13.0$ , $\rho_2= 19.0 g/cm^3$

$19.0 = 2.5 + C(13)$

=> $C = 1.27$

Now

$m = 8 \int\limits^{13}_{0} {2.5 + 1.27x} \, dx$

$m = 8 [{2.5 +\frac{ 1.27x^2}{2} } ]\left | 13} \atop {0}} \right.$

$m = 8 [{2.5 +\frac{ 1.27(13)^2}{2} } ]$

$m = 3454.08 g$

$m = 3.45408 kg$

8 0

2 years ago

An aluminum clock pendulum having a period of 1.00 s keeps perfect time at 20.0°C.

amm1812

Answer:

a) T ’= 0.999 s , b) t = 3596.4 s

Explanation:

The angular velocity of a simple pendulum is

w = √g / L

The angular velocity, frequency and period are related

w = 2π f = 2π / T

2π / T = √ g / L

T = 2π √ L / g

L = T² g / 4π²

L = 1² 9.8 / 4π²

L = 0.248 m

To know the effect of the temperature change let's use the thermal expansion ratios

ΔL = α L ΔT

ΔL = 24 10⁻⁶ 0.248 (-4 - 20)

ΔL = 142.8 10⁻⁶ m

Lf - L = -142. 8 10⁻⁶

Lf = 142.8 10⁻⁶ + 0.248

Lf = 0.2479 m

Let's calculate new period

T ’= 2π √ L / g

T ’= 2π √ (0.2479 / 9.8)

T ’= 0.999 s

We can see that the value of the period is reduced so that the clock is delayed

b) change of time in 1 hour

When the clock is at 20 ° C in one hour it performs 3600 oscillations, for the new period the time of this number of oscillations is

t = 3600 0.999

t = 3596.4 s

Therefore the clock is delayed almost 4 s

6 0

3 years ago

una onda longitudinal tiene una frecuencia de 200 hz y una longitud de onda de 4.2m ¿cual es la rapidez de la onda?​

una onda longitudinal tiene una frecuencia de 200 hz y una longitud de onda de 4.2m ¿cual es la rapidez de la onda?