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4 years ago

Geologists cannot yet predict earthquakes because

1 answer:

katrin2010 [14]4 years ago

3 0

<span>Geologists cannot yet predict earthquakes because the earthquakes caused by the movement of plates do not have the precise time and place when will it moves.</span>

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URGENTE ¿Cuál de los siguientes términos corresponde al concepto presión? * A. Es la fuerza perpendicular que ejerce un cuerpo s

melomori [17]

Answer:

Respuesta correcta, opción D: Es la fuerza que un cuerpo ejerce perpendicularmente sobre el área en la que actúa.

Explanation:

La definición de presión es la fuerza que un cuerpo ejerce en dirección perpendicular sobre el área en la que actúa.

Cuando se aplica una fuerza sobre la superficie de un cuerpo, la presión es la siguiente:

$P = \frac{F}{A}$

En donde:

F es la fuerza aplicada.

A es el área del cuerpo.

Por lo tanto la opción correcta es la D: es la fuerza que un cuerpo ejerce perpendicularmente sobre el área en la que actúa.

Espero que se sea de utilidad!

6 0

3 years ago

The two particles are both moving to the right. Particle 1 catches up with particle 2 and collides with it. The particles stick

Travka [436]

Answer:

c. vf is greator than v2, but less than v1

Explanation:

The principle of conservation of linear momentum states that when two or more bodies act upon one another, their total momentum remains constant.

In a system of colliding bodies the total momentum of the system just before the collision is the same as the total momentum just after the collision.

Collisions in which the kinetic energy is conserved are called elastic collision.

Collisions in which the kinetic energy is not conserved are called inelastic collisions. If the two objects stick together after the collision and move with a common velocity, the collision is said to be perfectly inelastic.

<em>The above scenario is a perfectly inelastic collision. The initial velocity of particle 1 was greater than particle 2 before collision. After collision, its velocity will reduce to a final velocity vf as it transfers some of its kinetic energy to particle 2; whereas, the velocity of particle 2 will increase to a final velocity vf as it absorbs some of the kinetic energy of particle 1.</em>

Therefore,

a. vf = v2 is wrong because vf is greater than v2

b. vf is less than v2 is wrong because vf is greater than v2

c. vf is greater than v2, but less than v1 is correct.

d. vf = v1 is wrong because vf is less than v1

4 0

3 years ago

A turtle ambles leisurely, as turtles tend to do, when it moves from a location with position vector 1,=1.91 m and 1,=−2.73 m in

Elena-2011 [213]

Answer:

Components: 0.0057, -0.0068. Magnitude: 0.0089 m/s

Explanation:

The displacement in the x-direction is:

$d_x = 3.65-1.91=1.74 m$

While the displacement in the y-direction is:

$d_y = -4.79 -(-2.73)=-2.06 m$

The time taken is t = 304 s.

So the components of the average velocity are:

$v_x = \frac{d_x}{t}=\frac{1.74}{304}=0.0057 m/s$

$v_y = \frac{d_y}{t}=\frac{-2.06}{304}=-0.0068 m/s$

And the magnitude of the average velocity is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(0.0057)^2+(-0.0068)^2}=0.0089 m/s$

8 0

3 years ago

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

Determine the diameter of the largest circular hole that can be punched into a sheet of polystyrene 6 mm thick, knowing that the

Yakvenalex [24]

Answer:

43.41 mm

Explanation:

Given:

thickness of sheet, t = 6 mm

Force exerted by punch, F = 45 KN

Average shearing stress, T = 55 MPa

From average shearing stress T = Force F / Area A

Hence area = force/stress =45000/ 55 = $818.18 mm^2$

From area = pi*diameter*thickness

diameter = area/(pi* thickness)

= 818.18/(3.142*6)

= 43.41 mm

4 0

3 years ago