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AnnZ [28]
3 years ago
14

Marco is conducting an experiment. He knows the wave that he is working with has a wavelength of 32.4 cm. If he measures the fre

quency as 3 hertz, which statement about the wave is accurate?
Physics
1 answer:
vekshin13 years ago
8 0

This question is incomplete; here is the complete question:

Marco is conducting an experiment. He knows the wave that he is working with has a wavelength of 32.4 cm. If he measures the frequency as 3 hertz, which statement about the wave is accurate?

A. The wave has traveled 32.4 cm in 3 seconds.

B. The wave has traveled 32.4 cm in 9 seconds.

C. The wave has traveled 97.2 cm in 3 seconds.

D. The wave has traveled 97.2 cm in 1 second.

The answer to this question is D. The wave has traveled 97.2 cm in 1 second.

Explanation:

The frequency of a wave, which is in this case 3 hertz, represents the number of waves that go through a point during 1 second. According to this, if the frequency of the wave is 3 hertz this means in 1 second there were 3 waves. Moreover, if you multiply the wavelength (32.4cm) by the frequency (3) you will know the distance the wave traveled in 1 second: 32.4 x 3 =  97.2 cm. This makes option D the correct one as the distance in 1 second was 97.2 cm.

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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

3 0
4 years ago
An object is 45.0 kg here on Earth. What is its mass on Jupiter if the acceleration
puteri [66]

Answer:

B. 45.0 kg

Explanation:

On earth the object has ;

Mass = 45 kg

acceleration = 9.81 m/s²

On Jupiter, acceleration is  24.79 m/s²

The mass of this object on Jupiter will be 45 kg. It will not change. Mass of an object will only change when you remove part of the object from it or add to it another part. The mass is the same on Earth and on Jupiter. However, due to increased acceleration on Jupiter , the weight will change/ increase because;

Weight = mass * acceleration

<u>On Earth </u>

Weight of the object will be : 45 *  9.81 = 441.45 kg

<u>On Jupiter</u>

Weight of the object will be : 45*24.79 =1115.55 kg

8 0
3 years ago
Sobre un barco, que se mueve en forma rectilínea, y con velocidad constante de 30 [km/h], se mueve un perro en el mismo sentido
almond37 [142]

Answer:

El observador verá correr al perro sobre la cubierta del barco a una velocidad de 40 kilómetros por hora.

Explanation:

Para determinar la velocidad del perro con respecto al observador sentado desde la playa a través del concepto de velocidad relativa, descrito en la siguiente fórmula:

v_{P/B} = v_{P} - v_{B} (1)

Donde:

v_{P/B} - Velocidad del perro relativo al barco, en kilómetros por hora.

v_{P} - Velocidad del perro con respecto al observador, en kilómetros por hora.

v_{B} - Velocidad del barco con respecto al observador, en kilómetros por hora.

Si sabemos que v_{B} = 30\,\frac{km}{h} y v_{P/B} = 10\,\frac{km}{h}, entonces la velocidad del perro con respecto al observador es:

v_{P} = v_{B} + v_{P/B}

v_{P} = 30\,\frac{km}{h} + 10\,\frac{km}{h}

v_{P} = 40\,\frac{km}{h}

El observador verá correr al perro sobre la cubierta del barco a una velocidad de 40 kilómetros por hora.

6 0
3 years ago
Question 10 of 10
soldier1979 [14.2K]
D. Energy that is transformed is neither destroyed nor created
5 0
3 years ago
Read 2 more answers
Help pleasee
agasfer [191]

Answer:

i/f = i/o + i/i       f = focal, o = object, i = image

1 / i = 1 / f - 1 / o  =    (o - f) / o f

i = o * f / ( o - f)      image distance

i = 12.5 * 22 / (12.5 - 22) = -28.9 cm

Image is real

Image is 28.9 cm to left of lens

M = - i / o = = 28.9 / 12.5 = 2.3     magnification (convex lens)

8 0
3 years ago
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