stone thrown vertically upward with speed of 15.5m/s from edge of cliff 75.0m high. how much later does it reach bottom of cliff?

Answers

Answer 1

Answer:

Answer:

Approximately $5.66\; {\rm s}$ , assuming that $g = 9.81\; {\rm m\cdot s^(-2)}$ and that air resistance is negligible.

Explanation:

Let upward be the positive direction. Under the assumptions, acceleration of the stone would be $a = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}$ (negative since the stone is accelerating downward.)

The duration of the flight can be found in the following steps:

Find velocity right before landing given displacement, initial velocity, and acceleration.
Find duration of the flight from acceleration and the change in velocity.

In SUVAT equation $v^(2) - u^(2) = 2\, a\, x$ :

$v$ is the final velocity right before landing (needs to be found,)
$u = 15.5\; {\rm m\cdot s^(-1)}$ is the initial velocity,
$a = (-9.81)\; {\rm m\cdot s^(-2)}$ is acceleration, and
$x = (-75.0)\; {\rm m}$ is displacement (downward because the stone landed below where it was launched.)

Rearrange this equation to find $v$ :

$\begin{aligned}v &= \sqrt{u^(2) + 2\, a\, x} \n &= \sqrt{(15.5)^(2) + 2\, (-9.81)\, (-75.0)} \; {\rm m\cdot s^(-1)} \n &\approx -40.05\; {\rm m\cdot s^(-1)}\end{aligned}$ .

In other words, the velocity of this stone has changed from the initial value of $u = (15.5)\; {\rm m\cdot s^(-1)}$ to $v \approx (-40.05)\; {\rm m\cdot s^(-1)}$ during the flight. Divide the change in velocity by acceleration $a = (-9.81)\; {\rm m\cdot s^(-2)}$ (the rate of change in velocity) to find the duration of the flight:

$\begin{aligned}t &= (v - u)/(a) \n &\approx ((-40.05) - (15.5))/((-9.81))\; {\rm s} \n &\approx 5.66\; {\rm s}\end{aligned}$ .

In other words, the stone would be in the air for approximately $5.66\; {\rm s}$ .

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During the motion of the slinky in a transverse wave, what do the particles of the slinky coil do?Question 3 options:

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the slinky particles move side to side

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Answers

The characteristics of the traveling waves we can find the correct answer for the movement of the particles in the transverse wave is:

The slinky particles move up and down.

The traveling wave movement is a way of transmitting energy without mass movement, it is formed by two movements united, an oscillatory movement and a displacement movement, there are two possibilities:

Longitudinal wave. In this case, the oscillation of the particles is in the same direction as the movement of the wave.

Transversal wave. The oscillation of the particle is perpendicular to the movement of the wave, see attached.

In both cases, the matter particles are oscillating around their equilibriumposition and the wave is the one that has a displacement movement.

Let's review the claims:

a) False. The particle has an oscillatory motion, it does not have a net displacement

b) False. The movement is oscillatory

c) True. As the wave is transversal, the oscillatory movement from top to bottom.

In conclusion using the characteristics of the traveling waves we can find the correct answer for the motion of the particles in the transverse wave is:

the slinky particles move up and down

Learn more here: brainly.com/question/14106293

Answer:

C.) The slinky particles move up and down

Explanation:

Transverse Wave-

A wave that has a disturbance perpendicular to the wave motion

Hello! This is the correct answer! Have a blessed day! :)

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Answers

Answer:

[See Below]

Explanation:

Your average speed would be 204.35 mph.

i got appx. 204.3478 miles per hour because you convert 7564km/23hours to 4700miles/23hours, then divide 4700 and 23 by 23 to end up with 204.3478miles/1hour

Suppose an object starts out electrically neutral. Through some process, 11 electrons are removed from the object. What is the electric charge of the object afterward? ANSWER 1. Unselected It has a net charge somewhere between the charge of 11 electrons and 11 protons, but we can’t tell exactly how much.
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3. Unselected It has the same net charge as 11 protons.
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Answers

Answer:

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Explanation:

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Answers

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Answers

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