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Natali [406]
3 years ago
9

A bug on the surface of a pond is observed to move up and down a total vertical distance of 6.5 cm , from the lowest to the high

est point, as a wave passes. If the ripples decreaseto 4.7 cm, by what factor does thebug's maximum KE change?
Physics
1 answer:
m_a_m_a [10]3 years ago
5 0

Answer:

factor that bug maximum KE change is 0.52284

Explanation:

given data

vertical distance = 6.5 cm

ripples decrease to =  4.7 cm

solution

We apply here formula for the KE of particle that executes the simple harmonic motion that is express as

KE = (0.5) × m × A² × ω²     .................1

and kinetic energy is  directly proportional to square of the amplitude.

so

\frac{KE2}{KE1} =  \frac{A2^2}{A1^2}      .............2

\frac{KE2}{KE1} = \frac{4.7^2}{6.5^2}

\frac{KE2}{KE1} = 0.52284

so factor that bug maximum KE change is 0.52284

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A 31.7 kg kid initially at rest slides down a frictionless water slide at 53.2 degrees, how fast is she moving in 3.45 s later?
Murrr4er [49]

Answer:

34.55 m/s

Explanation:

8 0
2 years ago
An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi
Veronika [31]

Answer:

s=5.79\ km

\theta=47^{\circ} east of south

Explanation:

Given:

  • distance of the person form the initial position, d'=8.4\ km
  • direction of the person from the initial position, 47^{\circ} north of east
  • distance supposed to travel form the initial position, d=5.3\ km
  • direction supposed to travel from the initial position, due North

<u>Now refer the schematic for visualization of situation:</u>

y=d'.\sin47^{\circ}-d

y=8.4\times \sin47-5.3 ...............(1)

x=d'.\cos47^{\circ}

x=8.4\times \cos47^{\circ} .................(2)

<u>Now the direction of the desired position with respect to south:</u>

\tan\theta=\frac{y}{x}

\tan\theta=\frac{8.4\times \sin47}{8.4\times \cos47}

\theta=47^{\circ} east of south

<u>Now the distance from the current position to the desired position:</u>

s=\sqrt{x^2+y^2}

s=\sqrt{(8.4\times \cos47)^2+(8.4\times \sin47-5.3)^2}

s=5.79\ km

4 0
3 years ago
How do very high density objects appear in an ultrasound?
evablogger [386]
<span>Density is entirely unrelated to an object's size. It is a property of a given</span>
7 0
3 years ago
A horizontal 826 N merry-go-round of radius 1.17 m is started from rest by a constant horizontal force of 57.8 N applied tangent
Julli [10]

Answer:

The kinetic energy of the merry-go-round is \bf{475.47~J}.

Explanation:

Given:

Weight of the merry-go-round, W_{g} = 826~N

Radius of the merry-go-round, r = 1.17~m

the force on the merry-go-round, F = 57.8~N

Acceleration due to gravity, g= 9.8~m.s^{-2}

Time given, t=3.47~s

Mass of the merry-go-round is given by

m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg

Moment of inertial of the merry-go-round is given by

I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}

Torque on the merry-go-round is given by

\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m

The angular acceleration is given by

\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}

The angular velocity is given by

\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}

The kinetic energy of the merry-go-round is given by

E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J

5 0
3 years ago
How will the amount of power change if less work is done in more time?
yKpoI14uk [10]

The amount of power change if less work is done in more time"then the amount of power will decrease".

<u>Option: B</u>

<u>Explanation:</u>

The rate of performing any work or activity by transferring amount of energy per unit time is understood as power. The unit of power is watt

Power = \frac{Work}{Time}  

Here this equation showcase that power is directly proportional to the work but dependent upon time as time is inversely proportional to the power i.e as time increases power decreases and vice versa.

This can be understood from an instance, on moving a load up a flight of stairs, the similar amount of work is done, no matter how heavy but  when the work is done in a shorter period of time more power is required.

7 0
3 years ago
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