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bezimeni [28]
3 years ago
9

Yo-Yo man releases a yo-yo from rest and allows it to drop, as he keeps the top end of the string stationary. The mass of the yo

-yo is 0.056 kg, its moment of inertia is 2.9×10−5kg⋅m2 and the radius, r, of the axle the string wraps around is 0.0064 m.
Through what height must the yo-yo fall for its linear speed to be 0.75 m/s ?
Physics
1 answer:
patriot [66]3 years ago
6 0

To solve this problem it is necessary to use the conservation equations of both kinetic, rotational and potential energy.

By definition we know that

KE + KR = PE

Where,

KE =Kinetic Energy

KR = Rotational Kinetic Energy

PE = Potential Energy

In this way

\frac{1}{2} mv^2 +\frac{1}{2} I\omega^2 = mgh

Where,

m = mass

v= Velocity

I = Moment of Inertia

\omega = Angular velocity

g = Gravity

h = Height

We know as well that \omega = v/r for velocity (v) and Radius (r)

Therefore replacing we have

\frac{1}{2} mv^2 +\frac{1}{2} I\omega^2 = mgh

[tex]h= \frac{1}{2} \frac{v^2}{g} +\frac{1}{2} \frac{I}{mg}(\frac{v}{r})^2[/tex]

h= \frac{1}{2}v^2 ( \frac{1}{g} +\frac{I}{mg}\frac{1}{r^2} )

h= \frac{1}{2}0.75^2 ( \frac{1}{9.8} +\frac{2.9*10^{-5}}{(0.056)(9.8)}\frac{1}{(0.0064)^2} )

h = 0.3915m

Therefore the height must be 0.3915 for the yo-yo fall has a linear speed of 0.75m/s

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creativ13 [48]

Given:

u(initial velocity)=0

a=5.54m/s^2

v(final velocity)=2 m/s

v=u +at

Where v is the final velocity.

u is the initial velocity

a is the acceleration.

t is the time

2=0+5.54t

t=2/5.54

t=0.36 sec


6 0
3 years ago
Two astronauts, each having a mass of 74.3 kg are connected by a 13.1 m rope of negligible mass. They are isolated in space, orb
murzikaleks [220]

Answer:

  L = 5076.5 kg m² / s

Explanation:

The angular momentum of a particle is given by

         L = r xp

         L = r m v sin θ

the bold are vectors, where the angle is between the position vector and the velocity, in this case it is 90º therefore the sine is 1

as we have two bodies

       L = 2 r m v

let's find the distance from the center of mass, let's place a reference frame on one of the masses

        x_{cm} = \frac{1}{M} \sum  x_{i} m_{i}i

        x_{cm} = \frac{1}{m+m} ( 0 + l m)

        x_{cm} = \frac{1}{2m}  lm

        x_{cm} = \frac{1}{2}

        x_{cm} = 13.1 / 2 = 6.05 m

let's calculate

          L = 2  6.05  74.3  5.65

          L = 5076.5 kg m² / s

4 0
3 years ago
Read 2 more answers
A rock is dropped from the edge of a cliff into a pool of water. Assume free-fall acceleration is 10 m/s per second, and air res
il63 [147K]

Answer:

20 m

Explanation:

From the equation of motion,

S = ut+1/2gt²................................. Equation 1

Where S = Height, u = initial velocity, t = time, g = acceleration due to gravity.

Note: Because the rocked is being dropped from a height, acceleration due to gravity is positive (g), and initial velocity (u) is negative

Given: t = 2.0 s, g = 10 m/s², u = 0 m/s (dropped from height)

Substituting into equation 1

S = 0(2) + 1/2(10)(2)²

S = 5(4)

S = 20 m

Hence the height of the the cliff above the pool is 20 m

7 0
3 years ago
Objects can be charged by the transfer of electrons.<br> True<br> False
Sveta_85 [38]

Answer:

True

Explanation:

Whenever electrons are transferred between objects, neutral matter becomes charged. For example, when atoms lose or gain electrons they become charged particles called ions. Three ways electrons can be transferred are conduction, friction, and polarization. In each case, the total charge remains the same.

I tried, hope this helps :)

* I might be wrong though

4 0
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Meiosis is the correct answer because of the fact all the other answers are no where near correct
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