D, a<span>s a sound source moves away from a stationary object, the sound waves. It is the opposite of what happens. </span>
Answer:
λ = 28,14 m
Explanation:
To find the wavelength of the wave you use the following formula:
(1)
v: speed of the wave = 1,97 m/s
λ: wavelength
f: frequency of the wave = 0,07 Hz
You replace the values of v and f in the equation (1) and solve for λ:

hence, the wavelength of the wave is 28,14 m
The variation of water depth at spreading centers (ridges) controlled by isostasy as convective cooling cools the rocks much more effectively the than heat conduction.
<h3>What is convective heat transfer?</h3>
When heat transfer takes place between the two fluids in direct or indirect contact.
The lithosphere cools when it moves away from the ridge axis by sea floor spreading. The cooler rocks have low density, so the sea floor gets deeper as the lithosphere gets more dense.
Thus, the convective cooling cools the rocks much more effectively the than heat conduction.
Learn more about convective heat transfer
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Answer:
Moc = -613.25 [lb*in]
Explanation:
Este problema se puede resolver mediante la mecánica vectorial, es decir se realizara un analisis de vectores.
Primero se calculara el momento de la fuerza F_AB con respecto al punto O, debemos recordar que el momento con respecto a un punto se define como el producto cruz de la distancia por la fuerza.
(producto cruz)
Necesitamos identificar los puntos:
O (0,0,0) [in]
A (12,0,0) [in]
B (0, 24,8) [in]
C (12,24,0) [in]
![r_{A/O}=(12,0,0) - (0,0,0)\\r_{A/O} = 12 i + 0j+0k [in]\\AB = (0,24,8) - (12,0,0)\\AB = -12i+24j+8k [in]\\[LAB]=\frac{-12i+24j+8k}{\sqrt{(12)^{2} +(24)^{2} +(8)^{2} } }\\ LAB=-\frac{3}{7} i+\frac{6}{7}j+\frac{2}{7}k](https://tex.z-dn.net/?f=r_%7BA%2FO%7D%3D%2812%2C0%2C0%29%20-%20%280%2C0%2C0%29%5C%5Cr_%7BA%2FO%7D%20%3D%2012%20i%20%2B%200j%2B0k%20%5Bin%5D%5C%5CAB%20%3D%20%280%2C24%2C8%29%20-%20%2812%2C0%2C0%29%5C%5CAB%20%3D%20-12i%2B24j%2B8k%20%5Bin%5D%5C%5C%5BLAB%5D%3D%5Cfrac%7B-12i%2B24j%2B8k%7D%7B%5Csqrt%7B%2812%29%5E%7B2%7D%20%2B%2824%29%5E%7B2%7D%20%2B%288%29%5E%7B2%7D%20%7D%20%7D%5C%5C%20LAB%3D-%5Cfrac%7B3%7D%7B7%7D%20i%2B%5Cfrac%7B6%7D%7B7%7Dj%2B%5Cfrac%7B2%7D%7B7%7Dk)
El ultimo vector calculado corresponde al vector unitario (magnitud = 1) de AB. El vector fuerza corresponderá al producto del vector unitario por la magnitud de la fuerza = 200 [lb].
![F_{AB}=-\frac{600}{7} i +\frac{1200}{7}j+\frac{400}{7} k [Lb]](https://tex.z-dn.net/?f=F_%7BAB%7D%3D-%5Cfrac%7B600%7D%7B7%7D%20i%20%2B%5Cfrac%7B1200%7D%7B7%7Dj%2B%5Cfrac%7B400%7D%7B7%7D%20k%20%5BLb%5D)
De esta manera realizando el producto cruz tenemos

![M_{O}=0i-685.7j+2057.1k [Lb*in]](https://tex.z-dn.net/?f=M_%7BO%7D%3D0i-685.7j%2B2057.1k%20%5BLb%2Ain%5D)
Para calcular el momento con respecto a la diagonal OC, necesitamos el vector unitario de esta diagonal.
![OC = (12,24,0)-(0,0,0)\\OC= 12i+24j+0k[Lb]\\LOC = \frac{12i+24j+0k}{\sqrt{(12)^{2} +(24)^{2} +(0)^{2} } } \\LOC=\frac{12}{\sqrt{720}}i+\frac{24}{\sqrt{720}}j +0k](https://tex.z-dn.net/?f=OC%20%3D%20%2812%2C24%2C0%29-%280%2C0%2C0%29%5C%5COC%3D%2012i%2B24j%2B0k%5BLb%5D%5C%5CLOC%20%3D%20%5Cfrac%7B12i%2B24j%2B0k%7D%7B%5Csqrt%7B%2812%29%5E%7B2%7D%20%2B%2824%29%5E%7B2%7D%20%2B%280%29%5E%7B2%7D%20%7D%20%7D%20%5C%5CLOC%3D%5Cfrac%7B12%7D%7B%5Csqrt%7B720%7D%7Di%2B%5Cfrac%7B24%7D%7B%5Csqrt%7B720%7D%7Dj%20%20%2B0k)
El vector con respecto al eje OC, es igual al producto punto del momento en el punto O por el vector unitario LOC
![M_{OC}=L_{OC}*M_{O}\\M_{OC}=(\frac{12}{\sqrt{720}}i +\frac{24}{\sqrt{720}} j+0k )* (0i-685.7j+2057.1k)\\M_{OC}= -613.32[Lb*in]](https://tex.z-dn.net/?f=M_%7BOC%7D%3DL_%7BOC%7D%2AM_%7BO%7D%5C%5CM_%7BOC%7D%3D%28%5Cfrac%7B12%7D%7B%5Csqrt%7B720%7D%7Di%20%2B%5Cfrac%7B24%7D%7B%5Csqrt%7B720%7D%7D%20j%2B0k%20%29%2A%20%280i-685.7j%2B2057.1k%29%5C%5CM_%7BOC%7D%3D%20-613.32%5BLb%2Ain%5D)
A would be the best answer to this question