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3 years ago

14

Two objects are dropped from different heights at the same instant. If the first object takes 10.7 seconds to hit the ground and

the second takes 14.8 seconds what was the difference in their original heights?

1 answer:

Lubov Fominskaja [6]3 years ago

5 0

Answer:

The answer to your question is : 521.8 m

Explanation:

Data:

Different heights

Time first object (tfo) = 10.7 s

Time second object (tso)= 14.8 s

Initial speed of both objects(vo) = 0 m/s

a = 9.81 m/s²

Formula:

h = vot + 1/2 (a)(t)² but vo = 0 so, h = 1/2 (a)(t)²

Then, height fo h = 1/2 (9.81)(10.7)² = 561.6 m

height so h = 1/2(9,81)(14.8)² = 1074.4 m

Difference in their heights = 1074.4 m - 561.6 m = 521.8 m

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A torque of magnitude T = 11 kN·m is applied to the end of a tank containing compressed air under a pressure of 8 MPa. Knowing t

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Max Normal Stress, $\sigma_{m} = 63.6\ MPa$

Max Shear Stress, $\tau_{m} = 34.3\ MPa$

Solution:

As per the question:

Torque, T = 11 kN.m = 11000 Nm

The inner diameter of the tank, $d_{i} = 180\ mm = 0.18\ m$

Thickness of the tank, t = 12 mm = 0.012 m

Pressure, P = 8 MPa = $8\times 10^{6}\ Pa$

Now,

Inner radius of the tank, $R_{i} = \frac{d_{i}}{2} = 0.09\ m$

Outer radius of the tank, $R_{o} = R_{i} + t = 0.9 + 0.012 = 0.102\ m$

Now,

To calculate the polar moment of inertia, J:

$J = \frac{\pi}{2}(R_{o}^{4} - R_{i}^{4})$

$J = \frac{\pi}{2}(0.102^{4} - 0.09^{4})$

$J = \frac{\pi}{2}(0.102^{4} - 0.09^{4}) = 6.7\times 10^{- 5}\ m^{4}$

Now,

To calculate the shear stress at the surface end:

$\sigma = \frac{P\times R_{i}}{t}$

$\sigma = \frac{8\times 10^{6}\times 0.09}{0.012} = 60\times 10^{6}\ Pa = 60\ MPa$

Now,

To calculate the stress at the outer end:

$\sigma' = \frac{P\times R_{i}}{2t}$

$\sigma' = \frac{8\times 10^{6}\times 0.09}{2\times 0.012} = 30\times 10^{6}\ Pa = 30\ MPa$

Now,

To calculate the value of the shear stress,

$\tau = \frac{TR_{o}}{J}$

$\tau = \frac{11000\times 0.102}{6.7\times 10^{- 5}} = 18.27\times 10^{6}\ Pa$

$\sigma = \sigma_{x}$

$\sigma' = \sigma_{y}$

$\tau_{xy} = \tau$

The resultant stress, $\sigma_{R} = 0.5(\sigma + sigma') = 0.5(90) = 45\ MPa$

The overall resultant of all the stresses (all in MPa):

$R = \sqrt{(\frac{\sigma - sigma'}{2})^{2} + \tau^{2}}$

$R = \sqrt{(\frac{60 - 30}{2})^{2} + 18.27^{2}} = 23.638\ MPa$

Now,

Max stress, $\sigma_{m} = R + \sigma_{R}$

$\sigma_{m} = 23.638 + 45 = 68.638\ MPa$

Minimum stress, $\sigma_{min} = 0\ MPa$

Now,

Maximum shear stress, $\tau_{m} = \frac{1}{2}(\sigma_{m} - \sigma_{min})$

$\tau_{m} = \frac{1}{2}(68.638 - 0) = 34.319\ MPa$

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