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Alexeev081 [22]
3 years ago
6

How do particles move when a transverse wave passes through a medium?

Physics
1 answer:
bagirrra123 [75]3 years ago
4 0
<span> In </span>transverse waves<span>, </span>particles<span> of the</span>medium<span> vibrate </span>to<span> and from in a direction perpendicular </span>to<span> the direction of energy transport. </span>
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3 years ago
Which of these actions requires KINETIC Energy? Choose all of the correct answers.
Keith_Richards [23]

Answer:

turning a doorknob

Explanation:

it will snap back once you release it.

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2 years ago
Read 2 more answers
Are basins a collection of smaller watersheds
mihalych1998 [28]

Explanation:

both are areas of land that drain to particular water bodies such as lakes

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3 years ago
A cylinder has a height of 5.3 cm and a diameter of 3.0 cm. what is its volume?
anzhelika [568]
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2 years ago
A brick lands 10.1 m from the base of a building. If it was given an initial velocity of 8.6 m/s [61º above the horizontal], how
Montano1993 [528]
<h2>Answer: 10.52m</h2><h2 />

First, we have to establish the <u>reference system</u>. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).

According to this, the initial velocity V_{o} has two components, because the brick was thrown at an angle \alpha=61\º:

V_{ox}=V_{o}cos\alpha   (1)

V_{ox}=8.6\frac{m}{s}cos(61\º)=4.169\frac{m}{s}  (2)

V_{oy}=V_{o}sin\alpha   (3)

V_{oy}=8.6\frac{m}{s}sin(61\º)=7.521\frac{m}{s}   (4)

As this is a projectile motion, we have two principal equations related:

<h2>In the x-axis: </h2>

X=V_{ox}.t  (5)

Where:

X=10.1m is the distance where the brick landed

t is the time in seconds

If we already know X and V_{ox}, we have to find the time (we will need it for the following equation):

t= \frac{X}{ V_{ox}}  (6)

t=2.42s  (7)

<h2>In the y-axis: </h2>

-y=V_{oy}.t+\frac{1}{2}g.t^{2}   (8)

Where:

y is the height of the building (<u>in this case it has a negative sign because of the reference system we chose)</u>

g=-9.8\frac{m}{s^{2}} is the acceleration due gravity

Substituting the known values, including the time we found on equation (7) in equation (8), we will find the height of the building:

-y=(7.521\frac{m}{s})(2.42s)+\frac{1}{2}(-9.8\frac{m}{s^{2}}).(2.42s)^{2}   (9)

-y=-10.52m   (10)

Multiplying by -1 each side of the equation:

y=10.52m >>>>This is the height of the building

3 0
3 years ago
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