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3 years ago

PLEASE HELP!!

Physics

2 answers:

malfutka [58]3 years ago

8 0

Answer:

Please see the explanation

Explanation:

As we know that standing waves have nodes and anti nodes which is produced by repeated interference of two waves of same frequency in opposite direction in the same medium. In this case when we have a rope attached to a point on a wall, we move the rope in up and down direction we will create waves.

The wave will go up to the wall and then reflect back and thus create an opposite wave. And as we can create a standing wave.

Arisa [49]3 years ago

3 0

Answer

standing waves are created when you have a rope attached to a point By moving your hand up and down we can create waves, and as we know that our rope is attached to a point on a wall, which means it has boundary. The waves will reflect back and interface with new waves and this is how standing waves are created.

hope this will help.

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The latitude of the city of Arlington is about 32.7357°. Calculate the following: (a) The angular speed of the Earth. (b) The li

julia-pushkina [17]

Answer:

Explanation:

a ) The earth rotates by 2π radian in 24 x 60 x 60 s

so angular speed ( w ) = 2π / (24 x 60 x 60) = 7.268 x 10⁻⁵ rad / s

b ) Linear speed of city of Arlington ( v ) = w r = w R Cosλ where R is radius of the earth and λ is latitude .

v = 7.268 x 10⁻⁵ x 6.371 x 10⁶ cos 32.7357

389.5 m /s

acceleration = w² r = w² R Cos 32.7357

= (7.268 x 10⁻⁵ )² x 6.371 x 10⁶ x cos 32.7357

=283.08 x 10⁻⁴ m/s²

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R cos 32.73/ R

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3 years ago

Question 7 (1 point)

ioda

Answer:

1150 secs

distance = speed x time

time = distance / speed

230,000 / 200 = 1150

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3 years ago

Alex pushes on a 2.0 kg book, resulting in a net force of 6.0 N on the book.

Yakvenalex [24]

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

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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 $\alpha$ :

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

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