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Virty [35]
3 years ago
5

A 0.30-m-radius automobile tire accelerates from rest at a constant 2.0 rad/s2. What is the centripetal acceleration of a point

on the outer edge of the tire after 5.0 s
Physics
1 answer:
lianna [129]3 years ago
8 0

Answer:

The centripetal acceleration is 30 m/s²

Explanation:

Given that,

Radius = 0.30 m

Acceleration = 2.0 rad/s²

Time = 5.0

We need to calculate the centripetal acceleration

Using formula of angular velocity

\omega=a\times t

Where,

a = acceleration

t = time

Put the value into the formula

\omega=2.0\times5.0

\omega=10\ rad/s

Using formula of centripetal acceleration

a_{c}=r\omega^2

a_{c}=0.30\times10^2

a_{c}=30\ m/s^2

Hence, The centripetal acceleration is 30 m/s²

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For heat transfer purposes, an egg can be considered to be a 5.5-cm-diameter sphere having the properties of water. An egg that
Harrizon [31]

Answer:

The time taken is   t = 40007 sec  

Explanation:

From the question we are told that

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    The initial temperature of egg the T_e = 4.3^{o}C

     The temperature of the boiling water T_b = 100^oC

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     The  thermal  conductivity of water is k = 0.607 W/m^oC

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Using one term approximation

We have the

            \frac{T_e_f - T_b}{T_e - T_b}  = Ae^{-\lambda ^2 \tau}

The radius is  r = \frac{5.5*10^ {-2}}{2} =2.75*10^{-2}m     Note that this radius is approximation to that of  a real egg

    Now we need to obtain the Biot number which help indicate the value of A  \ and \ \lambda to use in the above equation

     The Biot number is mathematically represented as

               Bi = \frac{H r}{k}

Substituting values  

               Bi = \frac{800 * 2.75 *10^{-2}}{0.607}

                    = 36.24

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        \lambda = 3.06632

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Substituting this into equation 1 we have

          \frac{74- 100}{4.3 - 100} = 1.9942 e^{-(3.0632^2) \tau}

          0.2717= 1.9942 e^{-(3.0632^2) \tau}

          0.2717= 1.9942 e^{-9.383 \tau}

           0.13624 =  e^{-9.383 \tau}

Taking natural log of both sides

           -1.993 =  -9.383\  \tau

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          t = \frac{\tau r^2}{\alpha }

substituting value  is  

         = \frac{0.2124 * 2.75 *10^{-2}}{0.146 *10^{-6}}

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8 0
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