How far away is a cliff if an echo is heard 0.486 s after the original sound? Assume that sound traveled at 343 m/s on that day.

Microphones and loudspeakers are used in an auditorium because the sound waves at the stage compared to the sound waves at the b

MaRussiya [10]

Acoustics (sounds)
Hope the this helps!

6 0

3 years ago

An electromagnet is strong enough to lift 200 kg of mass. If you double the number of turns of wire around the core and triple t

jekas [21]

Magnetic field is proportional to NIA.

N is the number of turns, I is the current and A is the area

You see N=2n, I = 3 i. Therefore, it will 6 times stronger

200*6= 1200 kg. It can lift now is 1200 kg.

6 0

3 years ago

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While playing her guitar , karen plucks one string with increasing levels of force. What effect does this have on the sound prod

Lena [83]

Answer:

The amplitude of vibration of string will increase due to which loudness of sound will increase

Explanation:

As we know that the guitar is based on the principle of Resonance. When string of the guitar vibrates at a given frequency then the sound produced in the hollow part of the guitar will also be at same frequency.

This is known as resonance condition, so guitar will produce same frequency sound as that of frequency of string.

Now if the string is plucked with increasing level of force then it will increase the amplitude of vibrations of the string due to which the sound produced in the guitar will also be of same level.

So here we can say that amplitude and intensity of sound related as

$I = kA^2$

so on increasing amplitude the intensity will increase and hence it will produce loud sound

8 0

3 years ago

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You are climbing in the high sierra when you suddenly find yourself at the edge of a fog-shrouded cliff. to find the height of t

sergey [27]

The total time it takes the sound to reach our on top of the cliff (10.0 s) is actually sum of two times:
- the time it takes the rock to hit the ground starting from the top of the cliff, t1
- the time it takes the sound to reach the top of the cliff starting from the ground, t2
1) The rock moves by uniformly accelerated motion, with constant acceleration $g=9.81 m/s^2$ . Its law of motion is given by $y(t)=h-v_0t-\frac{1}{2}gt^2$ , where $v_0=0$ is the initial velocity of the rock, and h is the height of the cliff. The time t1 is the time at which the rock reaches the ground, so that y(t1)=0, and the equation becomes
$0=h-\frac{1}{2}gt_1^2$
2) The sound moves from the ground to the top of the cliff by uniform motion, with constant speed v=343 m/s. Therefore, the sound covers the distance h (the height of the cliff) in a time t2 given by
$h=vt_2$
3) If we rewrite h in both equations, we can write:
$=\frac{1}{2}gt_1^2=vt_2$ (1)
4) We also know that the sum of t1 and t2 is equal to 10 seconds:
$t_1+t_2=10$
from which we find
$t_2=10-t_1$
if we substitute this into eq.(1), we get
$\frac{1}{2}gt_1^2+vt_1-10v=0$
Numerically:
$4.9t_1^2 + 343 t_1 - 3430=0$
Solving the equation, we find the solution $t_1=8.87 s$ (the other solution is negative, so it does not have physical meaning). As a consequence,
$t_2 = 10-t_1 = 1.13 s$
and the height of the cliff is given by
<span> $h=vt_2=(343 m/s)(1.13 s)=388 m$ </span>

3 0

3 years ago

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a tank with a flat bottom is filled with water to a height of 4 meters. what is the pressure at any point at the bottom of the t

Natasha_Volkova [10]

-- Each square meter of the bottom of the tank has 4 cubic meters
of water standing on it.

-- 4 cubic meters of water is equivalent to 4,000 liters.

-- So its mass is roughly 4,000 kilograms, and its weight is roughly
(4,000 x 9.81) = 39,240 newtons.

-- So the pressure on the bottom is roughly 39,240 newtons per
square meter, or 39,240 pascals, or <em>39.24 kPa</em>.

3 0

3 years ago

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