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3 years ago

10

If the total mass hanging on one end of the string that creates the tension in the string is 0.300 kg , what is the mass density

(mass per unit length) μ of the string?

1 answer:

soldi70 [24.7K]3 years ago

7 0

Answer:

The mass density of the string is (0.3/L)kg/m

Explanation:

Mass density of the string = Mass/Length

Mass = 0.3kg

The length of the string is unknown so it is assumed to be L meter(s)

Therefore, mass deny of the string = 0.3kg/Lm = (0.3/L)kg/m

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Answer:

a) $h=250\ m$

b) $\Delta h=0.0835\ m$

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Given:

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time after the takeoff after which the engine is shut off, $t_a=10\ s$

a)

<u>Maximum height reached by the helicopter:</u>

using the equation of motion,

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t = time of observation

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time after which Austin Powers deploys parachute(time of free fall), $t_f=7\ s$

acceleration after deploying the parachute, $a_p=2\ m.s^{-2}$

<u>height fallen freely by Austin:</u>

$h_f=u.t_f+\frac{1}{2} g.t_f^2$

where:

$u=$ initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)

$t_f=$ time of free fall

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$v_f=u+g.t_f$

$v_f=0+9.8\times 7$

$v_f=68.6\ m.s^{-1}$

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$h=u.t_h+\frac{1}{2} g.t_h^2$

where:

$u=$ initial velocity of the helicopter just before it begins falling freely = 0

$t_h=$ time taken by the helicopter to fall on ground

$h=$ height from where it falls = 250 m

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$250=0+0.5\times 9.8\times t_h^2$

$t_h=7.1429\ s$

From the above time 7 seconds are taken for free fall and the remaining time to fall with parachute.

<u>remaining time,</u>

$t'=t_h-t_f$

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$h'=v_f.t'+\frac{1}{2} a_p.t'^2$

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