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8_murik_8 [283]
3 years ago
6

A man releases a stone ar the top edge of a tower during the last second of its travel the stone falls through a distance of (9/

25)H where H is the towes highr. Find H
Physics
2 answers:
Lapatulllka [165]3 years ago
7 0

equations of motion (EoM) use EoM <span><span>v2</span>=<span>u2</span>+2ax</span> to establish velocities at positions shown in blue in drawing from EoM <span>v=u+at</span> for final 1 second of flight time, we can say <span>v=u+g(1)</span> <span><span><span>2gH</span><span>−−−−</span>√</span>=<span><span>2g<span>1625</span>H</span><span>−−−−−−</span>√</span>+g</span><span> then, solve for H [in terms of g]</span>
Musya8 [376]3 years ago
4 0
Equations of motion (EoM) use EoM <span>v2=u2+2ax</span> to establish velocities at positions shown in blue in drawing from EoM v=u+at for final 1 second of flight time, we can say v=u+g(1) <span><span>2gH−−−−√</span>=<span><span>2g1625H</span>−−−−−−√</span>+g</span><span> then, solve for H [in terms of g]

</span>
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Explanation:

For a particle under the action of a conservative force, its mechanical energy at point 0 is must be equal to its mechanical energy at point 1:

K_1+U_1=K_0+U_0\\\\\frac{mv_1^2}{2}+[(8.0\frac{J}{m^2})(x_1)^2+(2.0\frac{J}{m^4})(x_1)^4]=\frac{mv_0^2}{2}+[(8.0\frac{J}{m^2})(x_0)^2+(2.0\frac{J}{m^4})(x_0)^4]

In x_1=1.0m the speed is given, so v_1=5.0\frac{m}{s} and x_0=0. Replacing:

\frac{mv_1^2}{2}+[(8.0\frac{J}{m^2})(1m)^2+(2.0\frac{J}{m^4})(1m)^4]=\frac{mv_0^2}{2}+[(8.0\frac{J}{m^2})0^2+(2.0\frac{J}{m^4})(0)^4]\\\frac{mv_1^2}{2}+8.0J+2.0J=\frac{mv_0^2}{2}\\\frac{mv_0^2}{2}=\frac{mv_1^2}{2}+10J\\v_0=\sqrt{\frac{2}{m}(\frac{mv_1^2}{2}+10J)}\\v_0=\sqrt{\frac{2}{0.2kg}(\frac{(0.2kg)(5\frac{m}{s})^2}{2}+10J)}\\v_0=11.18\frac{m}{s}

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kenny6666 [7]

Answer: Within any frame of reference that is accelerating

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2. <em>The speed of light in vacuum has the same value for all inertial systems. </em>

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3 years ago
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