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3 years ago

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a rocket is launched with a constant acceleration straight up. exactly 4.00 seconds after lift off, a bolt falls off the side of

the rocket and hits the ground 6.00 seconds later. what was the rockets acceleration?

1 answer:

Papessa [141]3 years ago

6 0

The initial speed of the bolt is not 58.86 m/s.

Let a be the acceleration of the rocket.

During the 4 sec lift off, the rocket has reached a height of

h = (1/2)*a*t^2

with t=4,

h = (1/2)*a^16

h = 8*a

Its velocity at 4 sec is

v = t*a

v = 4*a

The initial velocity of the bolt is thus 4*a.

During the 6 sec fall, the bolt has the initial velocity V0=-4*a and it drops a total height of h=8*a. From the equation of motion,

h = (1/2)*g*t^2 + V0*t

Substituting h0=8*a, t=6 and V0=-4*a into it,

8*a = (1/2)*g*36 - 4*a*6

Solving for a

a = 5.52 m/s^2

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$\mu = \frac{8.79*10^{-3}}{70*10^{-2}}$

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The expression for the wavelength of the standing wave for the second overtone is

$\lambda = \frac{2}{3} l$

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Then,

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$f_1 = \frac{f_3}{3}$

$f_1 = \frac{448Hz}{3}$

$f_1 = 149.9Hz$

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