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shtirl [24]
3 years ago
15

A bicycle wheel is mounted on a fixed, frictionless axle, with a light string wound around its rim. The wheel has moment of iner

tia I=kmr2, where m is its mass, r is its radius, and k is a dimensionless constant between zero and one. The wheel is rotating counterclockwise with angular speed ω0_0, when at time t=0 someone starts pulling the string with a force of magnitude F. Assume that the string does not slip on the wheel.
The force F pulling the string is constant; therefore the magnitude of the angular acceleration α of the wheel is constant for this configuration.1. Find the magnitude of the angular velocity ω of the wheel when the string has been pulled a distance d.2. Express the angular velocity ω of the wheel in terms of the displacement d, the magnitude F of the applied force, and the moment of inertia of the wheel Iw.
Physics
1 answer:
Stells [14]3 years ago
3 0

Answer:

x3

Explanation:

trying to find the inetria magnitude of the wheel

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A concave lens has a focal length of -31 {cm}. Find the image distance and magnification that results when an object is placed 2
kolezko [41]

Answer:

The value is v  =  440 \  cm

and  

 m  =  16

Explanation:

From the question we  are told that

    The image  distance is  v  =  29 \  cm

    The focal length is f =  -31 \  cm

From the lens equation we have that

      \frac{1}{f}  =  \frac{1}{v}  - \frac{1}{u}

=>    \frac{1}{-31}  =  \frac{1}{v}  - \frac{1}{29}

=>     v  =  450 \  cm

Generally the magnification  is  

      m  =  \frac{v}{u}

=>    m  =  \frac{ 450}{29}

=>    m  =  16

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Nadusha1986 [10]

Answer:

the answer is c

Explanation:

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3 years ago
Weight is the amount of matter in an object :true or false
LekaFEV [45]

Answer: True

Explanation:

8 0
3 years ago
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Answer:

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4 years ago
A 60.0 kg ballet dancer stands on her toes during a performance with four square inches (26.0 cm2) in contact with the floor. Wh
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Answer:

Explanation:

Given

Mass of dancer m=60\ kg

Area of contact A=26\ cm^2

(a)If dancer is stationary

Pressure exerted by Floor is given by load per unit area of contact

P=\frac{weight}{A}

P=\frac{mg}{A}

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So apparent weight of dancer is W'=mg+ma

W'=m(g+a)=60\cdot (9.8+5)

W'=888\ N

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Pressure=\frac{888}{26\times 10^{-4}}

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5 0
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