Answer:
Just as distance and displacement have distinctly different meanings (despite their similarities), so do speed and velocity. Speed is a scalar quantity that refers to "how fast an object is moving." Speed can be thought of as the rate at which an object covers distance. A fast-moving object has a high speed and covers a relatively large distance in a short amount of time. Contrast this to a slow-moving object that has a low speed; it covers a relatively small amount of distance in the same amount of time. An object with no movement at all has a zero speed.
Answer: The right answer is b)
Explanation:
By definition, acceleration is the change in velocity (in module or direction) over a given time interval, as follows:
a = (v-v₀) / (t-t₀)
If we take t₀ = 0 (this is completely arbitrary), we can rewrite the equation above, as follows:
v = v₀ + at
We can recognize this function as a linear one, where a represents the slope of the line.
If a is constant, this means that the relationship between the change in velocity and the change in time remains constant, in other words, in equal times, its velocity changes in an equal amount.
Let's suppose that a = 10 m/s/s. (Usually written as 10 m/s²).
This is telling us that each second, the velocity increases 10 m/s.
I think this might be the answer let me know 0.2 g/cm^3
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)
