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3 years ago

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A hiker is at the bottom of a canyon facing the canyon wall closest to her. She is 280.5 meters from the wall and the sound of h

er voice travels at 340 m/s at that location. How long after she shouts will she hear her echo? (Be careful to consider why echoes happen.)

1 answer:

ValentinkaMS [17]3 years ago

8 0

Answer:

4.80 seconds

Explanation:

The velocity of sound is obtained from;

V= 2d/t

Where;

V= velocity of sound = 329.2 ms-1

d= distance from the wall = 790.5 m

t= time = the unknown

t= 2d/V

t= 2 × 790.5/ 329.2

t= 4.80 seconds

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A hungry 11.5 kg predator fish is coasting from west to east at 75.0 cm/s when it suddenly swallows a 1.25 kg fish swimming from

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Answer with Explanation:

We are given that

Mass,m=11.5 kg

$v_x=75 cm/s=0.75 m/s,v_y=0$

1 m=100 cm

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$v'_x=0,v'_y=3.6 m/s$

According to law of conservation momentum along east west direction

$11.5\times 0.75+1.25(0)=(11.5+1.25)V_x$

$8.625=12.75V_x$

$V_x=\frac{8.625}{12.75}$

$V_x=0.68 m/s$

According to law of conservation of momentum along north south direction

$11.5(0)+1.25(3.6)=(11.5+1.25)V_y$

$4.5=12.75V_y$

$V_y=\frac{4.5}{12.75}=0.35 m/s$

$V=\sqrt{V^2_x+V^2_y}$

$V=\sqrt{(0.68)^2+(0.35)^2}$

$V=0.76 m/s$

Direction, $\theta=tan^{-1}(\frac{V_y}{V_x})$

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3 years ago

Objects of equal mass are oscillating up and down in simple harmonic motion on two different vertical springs. The spring consta

Zanzabum

Answer:

$k_2=920\ N/m$

Explanation:

Given that,

The spring constant of spring 1, $k_1=230\ N/m$

The motion of the object on spring 1 has twice the amplitude as the motion of the object on spring 2, $A_1=2A_2$

As the magnitude of the maximum velocity is the same in each case, it means the maximum kinetic energy is same in each case. In other words, the total energy is same.

$\dfrac{1}{2}k_1A_1^2=\dfrac{1}{2}k_2A_2^2$

$k_1A_1^2=k_2A_2^2$

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3 years ago

a hunter 412.5m from a cliff moves a distance x towards the cliff and fires a gun. he hears the echo from the cliff after 2.2sec

Inessa [10]

Answer: 49.5 m

Explanation:

The speed of sound $s$ is given by a relation between the distance $d$ and the time $t$ :

$s=\frac{d}{t}$ (1)

Where:

$s=330 m/s$ is the speed of sound in air (taking into account this value may vary according to the medium the sound wave travels)

$d=412.5 m-x$ since we are told th hunter was initially 412.5 meters from the cliff and then moves a distance $x$ towards the cliff

$t=\frac{2.2 s}{2}=1.1 s$ Since the time given as data (2.2 s) is the time it takes to the sound wave to travel from the hunter's gun and then go back to the position where the hunter is after being reflected by the cliff

Having this information clarified, let's isolate $d$ and then find $x$ :

$d=st$ (2)

$412.5 m-x=(330 m/s)(1.1 s)$ (3)

Finding $x$ :

$x=49.5 m$ This is the distance at which the hunter is from the cliff.

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3 years ago