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Anvisha [2.4K]
3 years ago
15

A hunter is standing on flat ground between two vertical cliffs that are directly opposite one another. He is closer to one clif

f than the other. He fires a gun and, after a while, hears three echoes. The second echo arrives 1.05 s after the first, and the third echo arrives 0.827 s after the second. Assuming that the speed of sound is 343 m/s and that there are no reflections of sound from the ground, find the distance (in m) between the cliffs.

Physics
1 answer:
Ugo [173]3 years ago
7 0

Answer : distance, d = 463.7 m

Explanation :  

According to the given condition it can be assumed that the first echo would be heard from closest cliff. The second echo is from farther cliff and the third echo is from the reflection between the two cliffs.

Let the distance between the first cliff and the point of firing is x and y is the distance between the second cliff and the point of firing.

Then the first echo will travel 2x distance, second will travel 2y distance and third will travel 2x +2y.

So, using above data :

2y=v\times t_3

and

2x=v(t_2+t_3)

On solving :

y=\dfrac{343\times0.827}{2}

y=141.8\ m

x=\frac{343\times(1.05+0.827)}{2}

x=321.9\ m

x = 321.9 m and y = 141.8 m

Hence, total distance between two cliffs is d = 321.9 m + 141.8 m = 463.7 m

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Explanation:

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Since Force = mass × acceleration

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Cross multiply

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Since impulse = Ft

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The object change in velocity (v-u) = Ft/m from eqn 1

Going to the question;

a) Impulse = Force (F) × time(t)

Given force = 14N and time = 3seconds

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A Pitot-static probe is used to measure the speed of an aircraft flying at 3000 m. If the differential pressure reading is 3300
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Explanation:

given,

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Assuming the air flowing above air craft is in-compressible, irrotational and steady so, we can use Bernoulli's equation to solve the problem.

using Bernoulli's equation

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where ρ is the density of the air at 3000 m

          v= \sqrt{\dfrac{2 \times \Delta P}{\rho}}

          v= \sqrt{\dfrac{2 \times 3300}{0.909}}

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(a) 69.3 J

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W=Fd cos \theta

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\theta=30^{\circ} is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

\Delta K = W

where

\Delta K is the change in kinetic energy

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Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

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The change in kinetic energy of the sled can be rewritten as:

\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2 (1)

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We can calculate the variation of kinetic energy of the sled, \Delta K, after it has travelled for d=3 m. Using the work-energy theorem again, we find

\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J

And substituting into (1) and re-arrangin the equation, we find

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garik1379 [7]

Answer:

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Where the <em>negative sign indicates that it is pointing upwards.</em>

4 0
3 years ago
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