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Neporo4naja [7]
3 years ago
10

An archer fires an arrow, which produces a muffled "thwok" as it hits a target. If the archer hears the "thwok" exactly 1 s afte

r firing the arrow and the average speed of the arrow was 40 mis, what was the distance separating the archer and the target? Use 340 m/ s for the speed of sound.
Physics
1 answer:
aniked [119]3 years ago
6 0

Answer:

35,79 meters

Explanation:

So, we got an archer, and we got a target. Lets call the distance between this two d.

Now, the archer fires the arrow, that, in a time t_{arrow} travels the distance d with a speed v_{arrow} of 40 m/s and hits the target. We can see that the equation will be:

v_{arrow} * t_{arrow} = d\\ \\40 \frac{m}{s} * t_{arrow} = d

Immediately after this, the arrow produces a muffled sound, which will travel the distance d at  340 m/s in a time t_{sound}. Obtaining :

v_{sound} * t_{sound} = d\\ \\340 \frac{m}{s} * t_{sound} = d.

Finally, the sound reaches the archer, exactly 1 second after he fired the bow, so:

t_{arrow} + t _{sound} = 1 s.

This equation allows us to write:

t _{sound} = 1 s - t_{arrow}.

Plugging this  relationship in the distance equation for the sound:

340 \frac{m}{s} * t_{sound} = d \\ \\ 340 \frac{m}{s} * (1 s- t_{arrow}) = d.

Now, we can replace d from the first equation, and obtain:

40 \frac{m}{s} * t_{arrow} = d \\ 40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * (1 s- t_{arrow}).

Now, we can just work a little bit:

40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * 1 s - 340 \frac{m}{s} * t_{arrow} \\ \\ 40 \frac{m}{s} * t_{arrow} + 340 \frac{m}{s} * t_{arrow} = 340 m \\ \\ 380 \frac{m}{s} * t_{arrow} = 340 m \\ \\ t_{arrow} = \frac{340 m}{380 \frac{m}{s}} \\ \\ t_{arrow} = 0.8947 s.

Now, we can just plug this value into the first equation:

40 \frac{m}{s} * t_{arrow} = d

40 \frac{m}{s} * 340/380 s = 35,79 s = d

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vA 61.2-kg circus performer is fired from a cannon that is elevated at an angle of 57.8 ° above the horizontal. The cannon uses
dsp73

Answer:

The effective spring constant of the firing mechanism is 1808N/m.

Explanation:

First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:

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v_0=\frac{20.8m}{(2.60s)\cos57.8\°}\\\\v_0=15.0m/s

Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:

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