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3 years ago

Compare relative age with absolute age.

Physics

1 answer:

oee [108]3 years ago

8 0

Answer & Explanation:

Relative age is the age of a rock layer (or the fossils it contains) compared to other layers. It can be determined by looking at the position of rock layers. Absolute age is the numeric age of a layer of rocks or fossils.

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g initial angular velocity of 39.1 rad/s. It starts to slow down uniformly and comes to rest, making 76.8 revolutions during the

MrRa [10]

Answer:

Approximately $-1.58\; \rm rad \cdot s^{-2}$ .

Explanation:

This question suggests that the rotation of this object slows down "uniformly". Therefore, the angular acceleration of this object should be constant and smaller than zero.

This question does not provide any information about the time required for the rotation of this object to come to a stop. In linear motions with a constant acceleration, there's an SUVAT equation that does not involve time:

$v^2 - u^2 = 2\, a\, x$ ,

where

$v$ is the final velocity of the moving object,
$u$ is the initial velocity of the moving object,
$a$ is the (linear) acceleration of the moving object, and
$x$ is the (linear) displacement of the object while its velocity changed from $u$ to $v$ .

The angular analogue of that equation will be:

$(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ , where

$\omega(\text{final})$ and $\omega(\text{initial})$ are the initial and final angular velocity of the rotating object,
$\alpha$ is the angular acceleration of the moving object, and
$\theta$ is the angular displacement of the object while its angular velocity changed from $\omega(\text{initial})$ to $\omega(\text{final})$ .

For this object:

$\omega(\text{final}) = 0\; \rm rad\cdot s^{-1}$ , whereas
$\omega(\text{initial}) = 39.1\; \rm rad\cdot s^{-1}$ .

The question is asking for an angular acceleration with the unit $\rm rad \cdot s^{-1}$ . However, the angular displacement from the question is described with the number of revolutions. Convert that to radians:

$\begin{aligned}\theta &= 76.8\; \rm \text{revolution} \\ &= 76.8\;\text{revolution} \times 2\pi\; \rm rad \cdot \text{revolution}^{-1} \\ &= 153.6\pi\; \rm rad\end{aligned}$ .

Rearrange the equation $(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ and solve for $\alpha$ :

$\begin{aligned}\alpha &= \frac{(\omega(\text{final}))^2 - (\omega(\text{initial}))^2}{2\, \theta} \\ &= \frac{-\left(39.1\; \rm rad \cdot s^{-1}\right)^2}{2\times 153.6\pi\; \rm rad} \approx -1.58\; \rm rad \cdot s^{-1}\end{aligned}$ .

7 0

3 years ago

A plane mirror is placed to the right of an object. The image formed by the mirror will be a

guajiro [1.7K]

Answer:

A plane mirror is placed to the right of an object. The image formed by the mirror will be a virtual image that appears to be on the left of the mirror.

Explanation:

3 0

3 years ago

Read 2 more answers

Una pelota se desliza hacia arriba por una pendiente se halla inicialmente a 6m de la parte más baja de dicha pendiente y tiene

jolli1 [7]

Answer:

a) vprom = - 0.6 [m/s] b) ver explicacion c) vf = 10.8 [m/s]

Explanation:

Para poder solucionar este problema debemos completar el enunciado del problema así como la pregunta, realizando una búsqueda en Internet encontramos el enunciado completo, así como la pregunta.

"Una pelota que se desliza hacia arriba por una pendiente se halla inicialmente a 6 m de la parte más baja de dicha pendiente y tiene una velocidad de 4 m/s. Cinco segundos después se encuentra a 3 m de la parte más baja. Si suponemos una aceleración constante, "

¿cual fue la velocidad media?

¿Cuál es el significado de una velocidad media negativa?

¿Cuáles son la aceleración media y la velocidad final?

Para facilitar esta solución debemos realizar un esquema del movimiento de la pelota en el plano inclinado. En la imagen adjunta se puede ver un esquema del movimiento de la pelota sobre el plano inclinado en los diferentes tiempos mencionados.

¿cual fue la velocidad media?

La velocidad media se debe calcular utilizando la expresión de velocidad = espacio / tiempo.

$v_{prom}=x/t\\v_{prom}=(3-6)/5\\v_{prom}=-0.6 [m/s]$