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lina2011 [118]
4 years ago
13

A hunter points his rifle directly at a coconut that she wishes to shoot off a tree. It so happens that the coconut falls from t

he tree at the exact instant the hunter pulls the trigger. Consequently, A. The bullet passes above the coconut B. The bullet hits the coconut C. The bullet passes beneath the coconut
Physics
1 answer:
Advocard [28]4 years ago
8 0

Answer:

B. The bullet hits the coconut

Explanation:

The hunter fired a bullet and hit the coconut he wanted to drop from the tree. It may not seem so, since the coconut fell as soon as the hunter pulled the trigger, but a rifle bullet has a speed equal to 3,500 km / h, which is too fast for our eyes to be able to observe the moment when the bullet hit the coconut.

However, it would be quite a coincidence that exactly the coconut that the hunter aimed for fell naturally without the influence of the bullet. Because of this, we can say that the hunter hit the coconut.

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White footed mouse, than human.
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You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh
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To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

\theta = \frac{1.22 \lambda }{D}

Here,

D is diameter of the eye

D = \frac{1.22 (539nm)}{5.11 mm}

D= 1.287*10^{-4}m

The angle that relates the distance between the lights and the distance to the lamp is given by,

Sin\theta = \frac{d}{L}

For small angle, sin\theta = \theta

sin \theta = \frac{d}{L}

Here,

d = Distance between lights

L = Distance from eye to lamp

For small angle sin \theta = \theta

Therefore,

L = \frac{d}{sin\theta}

L = \frac{0.691m}{1.287*10^{-4}}

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Therefore the distance is 5.367km.

4 0
4 years ago
List 3 quantities of waves.
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8 0
4 years ago
Read 2 more answers
If a fish experiences an a of 9.8m/s2 over a period of 1.84s, what was the 4v?
777dan777 [17]

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6 0
3 years ago
The distance between adjacent nodes in a standing wave pattern in a length of string is 25.0 cm:A. What is the wavelength of wav
mina [271]

A) 50 cm

B) 10000 cm/s

Explanation

Step 1

A)

If you know the distance between nodes and antinodes then use this equation:

\begin{gathered} \frac{\lambda}{2}=D \\ \text{where}\lambda\text{ is the wavelength} \\ D\text{ is the distance betw}een\text{ nodes} \end{gathered}

then, let

D=\text{ 25 cm }

now, replace to find the wavelength

\begin{gathered} \frac{\lambda}{2}=25 \\ \text{Multiply both sides by 2} \\ \frac{\lambda}{2}\cdot2=25\cdot2 \\ \lambda=50\text{ Cm} \end{gathered}

so, the wavelength is

A) 50 cm

Step 2

The speed of a wave can be found using the equation

v=\lambda f

or velocity = wavelength x frequency,

then,let

\begin{gathered} \lambda=50\text{ cm} \\ f=200\text{ Hz} \end{gathered}

replace and evaluate

\begin{gathered} v=\lambda f \\ v=50\text{ cm }\cdot200\text{ HZ} \\ v=10000\text{ }\frac{\text{cm}}{s} \end{gathered}

so

B) 10000 cm/s

I hope this helps you

6 0
1 year ago
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