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aleksandrvk [35]
3 years ago
13

You are generating traveling waves on a stretched string by wiggling one end. If you suddenly begin to wiggle more rapidly witho

ut appreciably affecting the tension, you will cause the waves to move down the string a. faster than before.b. at the same speed as before.c. slower than before.
Physics
1 answer:
german3 years ago
6 0

Answer:

Option as B is correct At the same speed as before

Explanation:

As we know the relation between speed of the wave and tension in string

The speed of wave in stretched string

ν = \sqrt{\frac{T}{\mu} }  

speed of wave is the directly proportional to the square root of tension as mentioned in question tension of string is unaffected when in linear mass density is constant,  so we can say that the  speed of wave will  be the same  

Option as B is correct At the same speed as before  

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Una ola oceánica viaja a aproximadamente 1,97 m / s. Esto es 4 millas por hora. La frecuencia de las ondas es de aproximadamente
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Una fuerza F de 200 lb actúa a lo largo de AB, sobre la rampa mostrada. fuerza de F respecto del eje OC. Calcule el momento de f
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Answer:

Moc = -613.25 [lb*in]

Explanation:

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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.

M_{o}=r_{A/O} * F_{AB} (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

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]

De esta manera realizando el producto cruz tenemos

M_{O}=r_{A/O} * F_{AB}

M_{O}=0i-685.7j+2057.1k [Lb*in]

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

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]

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