How do we know that an object has accelerated?

A zebra drinks from a watering hole and is ambushed by a crocodile. What's more important and why: the zebra having a high maxim

elena55 [62]

Answer:

before this type of attack, high acceleration is the most important thing.

Explanation:

As the zebra is ambushed by the crocodile the most important thing is a quick reaction, in this attack the most likely is that the crocodile is in the water so it cannot run after the zebra.

Consequently, before this type of attack, high acceleration is the most important thing.

5 0

2 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

Alice climbs 100 meters directly up the face of a cliff to reach the summit. Her friend Peter hikes the long way around taking a

SOVA2 [1]

Answer:

Alice Distance = 100 meters

Peter's Distance = 3 km

Alice Displacement and Peter's displacement are both 100 meters upwards.

Explanation:

To solve this question, we have to first define distance and displacement.

Distance is simply the measurement of the sum of all paths travelled from one point to another while displacement is measurement of the shortest distance from initial point to final point.

Now, Alice and Peter are moving from the same point.

Alice distance travelled is 100 meters.

Also, her displacement will be 100 meters because it is the shortest distance to the summit of the cliff.

Now, for Peter, he decides to take a longer route which is 3 km in distance.

However, the shortest path which is the displacement is still 100 meters.

Thus, Peter's displacement is 100 meters.

4 0

3 years ago

[OU.04] The picture below shows two galaxies. What of these statements best describes a similarity between the two galaxies?

Vitek1552 [10]

Answer:

D.

Explanation: Both have shapes determined by gravitational forces.

7 0

3 years ago

Read 2 more answers

In a liquid with a density of 1050 kg/m3, longitudinal waves with a frequency of 450 Hz are found to have a wavelength of 7.90 m

GaryK [48]

Answer:jdhddj Nd

Explanation:xbxnbxx period

8 0

3 years ago