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Anon25 [30]
3 years ago
7

A man, facing a high wall, notices that an echo is heard 4.0 seconds after he makes a sharp sound. After walking 200m directly t

owards the cliff, an echo is heard after 3.4 seconds. Calculate (i) the speed of the sound (ii) how far he was from the wall when he made his second observation of the echo
Physics
1 answer:
cestrela7 [59]3 years ago
4 0

Answer:2.4

Explanation:basically 2.4 = 2.500 kg and then we add 3.4 seconds in all of them 4.0 + 3.4 = 2.5 ok

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When the Moon passes between Sun and Earth, the lunar shadow is seen as a solar eclipse on Earth. When Earth passes directly between Sun and Moon, its shadow creates a lunar eclipse. Lunar eclipses can only happen when the Moon is opposite the Sun in the sky, a monthly occurrence we know as a full Moon.
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Why are the stars that appear in the night sky not visible during the day?
xxTIMURxx [149]
Because the sun is not hitting the light on them and there not as bright like when we see them when it's dark.
6 0
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A 0.20-kg mass is oscillating on a spring over a horizontal frictionless surface. When it is at a displacement of 2.6 cm for equ
valentinak56 [21]

Explanation:

The given data is as follows.

                    mass = 0.20 kg

              displacement = 2.6 cm

              Kinetic energy = 1.4 J

       Spring potential energy = 2.2 J

Now, we will calculate the total energy present present as follows.

         Total energy = Kinetic energy + spring potential energy

                           = 1.4 J + 2.2 J

                            = 3.6 Joules

As maximum kinetic energy of the object will be equal to the total energy.

So,      K.E = Total energy

                = 3.6 J

Also, we know that

                  K.E = \frac{1}{2}mv^{2}_{m}

or,                   v = \sqrt{\frac{2K.E}{m}}

                        = \sqrt{2 \times 3.6 J}{0.2 kg}

                        = \sqrt{36}

                        = 6 m/s

thus, we can conclude that maximum speed of the mass during its oscillation is 6 m/s.

4 0
3 years ago
How an application of atmospheric device work?example siphon​
Vika [28.1K]

Answer:

A practical siphon, operating at typical atmospheric pressures and tube heights, works because gravity pulling down on the taller column of liquid leaves reduced pressure at the top of the siphon (formally, hydrostatic pressure when the liquid is not moving).

I hope it's helpful!

6 0
3 years ago
A radio signal has a frequency of 1.023 x 108 HZ. If the speed of the signal in air is 2.997 x 108m/s, what is the wavelength of
Sladkaya [172]

Answer:

2.93 m  (which agrees with answer "C" on the list)

Explanation:

Recall that the speed of the wave equals the product of the wave's length times its frequency. Therefore, the wavelength is going to be the quotient of the speed of the signal divided its frequency:

Wavelength = 2.997  10^8 / 1.023  10^8 =  2.93 m

5 0
3 years ago
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