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chubhunter [2.5K]

3 years ago

10

A 1200 kg aircraft going 30 m/s collides with a 2000 kg aircraft that is parked and they stick together after the collision and

are going 11.3 m/s after the collision. If they skid for 14.7 seconds before stopping, how far did they skid

1 answer:

Leno4ka [110]3 years ago

4 0

Answer:

83.055 m

Explanation:

According to the given scenario, the calculation of skid distance is shown below:-

$S = \frac{1}{2} \times (u + v) \times t$

Where

u = 11.3

v = 0

t = 14.7

Now placing these values to the above formula,

So,

$S = \frac{1}{2} \times (11.3 + 0) \times 14.7$

= 83.055 m

Therefore for computing the skid distance we simply applied the above formula i.e by considering the all items given in the question

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Fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.20 mm cos[(

bezimeni [28]

Answer

given,

y(x,t)= 2.20 mm cos[( 7.02 rad/m )x+( 743 rad/s )t]

length of the rope = 1.33 m

mass of the rope = 3.31 g

comparing the given equation from the general wave equation

y(x,t)= A cos[k x+ω t]

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now on comparing

a) Amplitude = 2.20 mm

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f = 118.25 Hz

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$k= \dfrac{2\pi}{\lambda}$

$\lambda= \dfrac{2\pi}{k}$

$\lambda= \dfrac{2\pi}{7.02}$

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d) speed

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$v = \dfrac{743}{7.02}$

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e) direction of the motion will be in negative x-direction

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$T = \dfrac{v^2\ m}{L}$

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3 years ago

An inattentive driver is traveling 18.0 m/s when he notices a red light ahead. his car is capable of decelerating at a rate of 3

ryzh [129]

Speed of the car given initially

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$d_1 = v* t$

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Now after applying brakes the distance covered by the car before it stops is given by kinematics equation

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here

vi = 18 m/s

vf = 0

a = - 3.35

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$0^2 - 18^2 = 2*(-3.35)(s)$

$s = 48.35 m$

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vladimir2022 [97]

Answer:

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Explanation:

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Nana76 [90]

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