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AleksAgata [21]
3 years ago
15

An astronomer at the equator measures the Doppler shift of sunlight at sunset. From this, she calculates that Earth's tangential

velocity at the equator is 465 m/s. The centripetal acceleration at the equator is 3.41 * 10^-2 m/s2. Use this data to calculate Earth's radius.
Physics
1 answer:
Ksivusya [100]3 years ago
7 0

Answer : Radius of Earth is 6,340 Km.

Explanation

It is given that,

Tangential velocity of Earth at the equator, v = 465 m/s

Centripetal acceleration at the equation is, a_c=3.41\times 10^{-2}\ m/s^2

We know that the relation between centripetal acceleration and tangential velocity is :

a_c=\dfrac{v^2}{r}..........(1)

Where,

v is tangential velocity

r is the radius

Putting all values in equation (1)

3.41\times 10^{-2}\ m/s^2=\dfrac{(465 m/s)^2}{r}

r=6340909\ m

or

r=6,340\ Km

The Earth's radius is 6,340 Km. Hence, this is the required solution.

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Answer:

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Explanation:

From the question we are told that

   The height above  firefighter safety net is H  = 14 \ m

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From the law of energy conservation

    KE_T + PE_T =  KE_B + PE_B

 Where KE_T is the kinetic energy of the person before jumping which equal to zero(because to kinetic energy at maximum height )

   and  PE_T is the potential energy of the before jumping  which is mathematically represented at

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and  KE_B is the kinetic energy of the person just before landing on the safety net  which is mathematically represented at

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and  PE_B is the potential energy of the person as he lands on the safety net which has a value of zero (because it is converted to kinetic energy )

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=>           v =  \sqrt{2 gH }

    substituting values

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Applying the equation o motion

             v_f =  v  + 2 a s

Now the final velocity is zero because the person comes to rest

      So

         0 = 16.57 + 2 * a * 1.8

            a =  - \frac{16.57^2 }{2 * 1.8}

            a =  - 76.27 m/s^2

         

         

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3 years ago
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