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Alekssandra [29.7K]
3 years ago
9

A girl running around a circular track of radius 200 m completes one

Physics
1 answer:
kondor19780726 [428]3 years ago
4 0

Answer:

v = 2 m/s

Explanation:

First, we find the angular velocity of the girl by using the following formula:

Angular\ Velocity =\omega= \frac{Angular\ Distance\ Covered}{Time\ Taken}\\\\\omega = \frac{1\ revolution}{10\ min}\frac{1\ min}{60\ s}\frac{2\pi\ rad}{1\ revolution}\\\\\omega = 0.01\ rad/s

Now, for the average velocity of the girl we can use the following formula:

Average\ Velocity = V = r\omega\\V = (200\ m)(0.01\ rad/s)\\

where,

r = radius = 200 m

therefore,

<u>v = 2 m/s</u>

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Answer:

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Explanation:

Remember that the substance is steam so it's water (H2O) and the initial conditions are P_{1} =1MPa, T_{1}=400^{0}C, m=0.6Kg andv_{2} =0.4v_{1} from a saturated water table and the initial conditions we can determine that the state phase is superheated (see Table 1 attached) because the T_{sat}=179.88^{0} C \leq T_{1} from the table 1 we get:v_{1} =0.30661(m^{3}/Kg). Now we have second conditions as: P_{2}=1(MPa), T_{2}=250^{0}C so from the same table we can see the state still superheated and we getv_{2}=0.23275(m^{3}/Kg), knowing that it's a isobaric process we can find the compression's work as:W_{b}=m*P(v_{2}-v_{1})=0.6*1000*(0.23275-0.30661)=-44.32(KJ) so the compressor's work is: 44.32(KJ). (b) Then the piston reaches the stop and there are two processes in this stage, so Process 1 is isobaric and:W_{1}=m*P*(v_{2}-v_{1}) =0.6*1000*(0.4*0.30661-0.30661)=-110.38(KJ) and the second process is isochoric:W_{2}=zero,nowW_{b}=W_{1}+ W_{2} =110.38+0=110.38(KJ). Finally to get the temperarure at the final state in part (b) we get:v_{2} =0.4v_{1} =0.4*0.30661=0.122644(m^{3}/Kg), P_{2}=500(KPa) from table 2 (see attached) we comparev_{f} andv_{g} at the saturated water table and find the following:v_{f}=0.001093(m^{3}/Kg), so we know that the final state phase is a satured mixture and we get the temperature at the final state as:T_{2} =T_{sat} =151.83^{0}C.

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Complete Question

(A) What is the maximum tension possible in a 1.00- millimeter-diameter nylon tennis racket string?

(B) If you want tighter strings, what do you do to prevent breakage: use thinner or thicker strings? Why? What causes strings to break when they are hit by the ball?

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Answer:

A

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B

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Explanation:

From the question we are told that

     The  diameter is  d =  1.00 \ mm  =  0.001 \  m  

       The  tensile strength of the nylon string is \sigma =  600 *10^{6} \  N/m^2

  Generally the radius is mathematically evaluated as

     r=  \frac{d}{2}

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=>     r =  0.0005 \  m

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=>   A =  3.142  *  (0.005)^2

=>    A =  7.855*10^{-7}\  m^2

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           T =  \sigma  *  A

=>          T =  600*10^{6} *  7.855*10^{-7}

=>         T = 471.3 \  N

Form the equation above  we  see that

        T  \  \alpha \  A

So if the tension is  increased to prevent breakage the thickness of the string is increased(i. e the cross-sectional  area )

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