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gavmur [86]
1 year ago
7

Suppose you fill two rubber balloons with air, suspend both of them from the same point, and let them hang down on strings of eq

ual length. You then rub each with wool or on your hair so that the balloons hang apart with a noticeable separation between them. Make order-of-magnitude estimates of (a) the force on each
Physics
1 answer:
andrew-mc [135]1 year ago
3 0

The force on each balloon is 2×10^−3 N.

Consider two balloons of diameter 0.200m each with a mass of 1.00g hanging apart with 0.0500m separation on the ends of string making angles of 10.0° with the vertical.

\sum F_{y} = Tcos10\textdegree - mg = 0\\\\T = \frac{mg}{cos10\textdegree } \\\\\sum F_{y} = Tsin10\textdegree - mg = 0\\

So,

F_{e}  = \frac{mg}{cos10\textdegree }sin10\textdegree  = mgtan10\textdegree \\\\= (0.00100kg)(9.8m/s^{2})tan10\textdegree \\\\F_{e} = 2 \times 10^{-3}N

A force is an influence that can change the motion of an object. A force can cause an object with mass to change its velocity (e.g. moving from a state of rest), i.e., to accelerate. Force can also be described intuitively as a push or a pull. A force has both magnitude and direction, making it a vector quantity. It is measured in the SI unit of newton (N).

Learn more about force here:

brainly.com/question/13191643

#SPJ4

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Assume that, when we walk, in addition to a fluctuating vertical force, we exert a periodic lateral force of amplitude 25 NN at
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Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told

   The amplitude of the lateral  force is  F = 25 \  N

   The frequency is   f = 1 \  Hz

   The mass of the bridge per unit length is  \mu  =  2000 \  kg /m

    The length of the central span is  d =  144 m

     The oscillation amplitude of the section  considered at the time considered is  A = 75 \ mm =  0.075 \  m

      The time taken for the undriven oscillation to decay to \frac{1}{e}  of its original value is  t = 6T

Generally the mass of the section considered is mathematically represented as

            m =  \mu  *  d

=>        m =  2000 * 144

=>        m =  288000 \ kg

Generally the oscillation amplitude of the section after a  time period  t is mathematically represented as

                 A(t) = A_o e^{-\frac{bt}{2m} }

Here b is the damping constant and the A_o is the amplitude of the section when it was undriven

So from the question  

               \frac{A_o}{e}  = A_o e^{-\frac{b6T}{2m} }

=>            \frac{1}{e}  =e^{-\frac{b6T}{2m} }

=>          e^{-1} =e^{-\frac{b6T}{2m} }

=>           -\frac{3T b}{m}  =  -1

=>         b  = \frac{m}{3T}

Generally the amplitude of the section considered is mathematically represented as

           A =  \frac{n * F }{ b *  2 \pi }

=>       A =  \frac{n * F }{ \frac{m}{3T}  *  2 \pi }

=>       n =  A  *  \frac{m}{3}  *  \frac{2\pi}{25}

=>       n = 0.075 *  \frac{288000}{3}  *  \frac{2* 3.142 }{25}

=>       n = 1810 \ people

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Explanation:

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