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3 years ago

8

What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 1

0 24 kg and radius of the earth is 6.4 × 10 6 m.)

1 answer:

Free_Kalibri [48]3 years ago

7 0

Answer:

Explanation:

just use the gravational force equation which is G x m of earth x m of object divided by r squared (which is radius of earth)

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Complete Question

The complete question is shown on the uploaded image

Answer:

The tension on the shank is $T =8391.6 N$

Explanation:

From the question we are told that

The strain on the strain on the head is $\Delta l = 0.1 mm/mm = \frac{0.1}{1000} = 0.1 *10^{-3} m/m$

The contact area is $A = 2.8 mm^2 = 2.8* (\frac{1}{1000} )^2 = 2.8*10^{-6} m^2$

Looking at the first diagram

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To find the stress at $\Delta l$ we use the interpolation method

$\frac{\sigma_{\Delta l} - \sigma_{0.0015} }{ \sigma _ {0.3} - \sigma_{0.0015} } = \frac{e_{\Delta l } - e_{0.0015}}{e_{0.3} - e_{ 0.0015}}$

Substituting values

$\frac{\sigma _{\Delta l} - 450}{600 - 450} = \frac{0.1 -0.0015}{0.3 - 0.0015}$

$\sigma _{\Delta l} -450 = 49.50$

$\sigma _{\Delta l} = 499.50 MPa$

Generally the force on each head is mathematically represented as

$F = \sigma_{\Delta l} * A$

Substituting values

$F = 499.50*10^{6} * 2.8*10^{-6}$

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$T = 6 * F$

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