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vivado [14]
3 years ago
11

Wind blowing across suspended power lines may cause the power lines to vibrate at their natural frequency. This often produces a

udible sound waves. This phenomenon, often called an Aeolian harp, is an example of:
A) diffraction
B) refraction
C) the Doppler effect
D) resonance
Physics
1 answer:
Lilit [14]3 years ago
7 0

Answer: D resonance

Explanation: Resonance is the effect produced when an object is forced to vibrate at its natural frequency

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If the entire population of Earth were transferred to the Moon, how far would the center of mass of the Earth-Moon-population sy
Genrish500 [490]

The question is incomplete. The complete question is :

If the entire population of Earth were transferred to the Moon, how far would the center of mass of the Earth-Moon population system move? Assume the population is 7 billion, the average human has a mass of 65 kg, and that the population is evenly distributed over both the Earth and the Moon. The mass of the Earth is 5.97×1024 kg and that of the Moon is 7.34×1022 kg. The radius of the Moon’s orbit is about 3.84×105 m.

Solution :

Given :

Mass of earth, $M_e = 5.97 \times 10^{24} \ kg$

Mass of moon, $M_m = 7.34 \times 10^{22} \ kg$

Mass of each human, $m_p =65 \ kg$

Therefore mass of total population, $M_p = 65  \times 7 \times 10^{9} \ kg$

                                                           $M_p = 4.55 \times 10^{11} \ kg$

Let the earth is at the origin of the coordinate system. Then,

Since $M_e>> M_p$

         $M_m>> M_p$

Hence if we shift all the population on the moon there will be negligible change in the mass of the moon and earth. Hence there will not be any significant shift on the centre of mass. i.e.

      $X_{cm} = \frac{5.97 \times 10^{24}+ 7.34 \times 10^{22} \times 3.84 \times 10^5}{5.97 \times 10^{24}+ 7.34 \times 10^{22}}$

              $= 4.68 \times 10^6 \ m$

$ 4.68 \times 10^3 \ km$ from the earth.

         

3 0
3 years ago
A robot probe drops a camera off the rim of a 239 m high cliff on Mars, where the free-fall acceleration is -3.7 m/s^2. Find the
cestrela7 [59]

Answer:

42.05 m/s

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Height (h) = 239 m

Acceleration due to gravity (g) = 3.7 m/s²

Final velocity (v) =?

The velocity with which the camera hits the ground can be obtained as follow:

v² = u² + 2gh

v² = 0² + 2 × 3.7 × 239

v² = 0 + 1768.6

v² = 1768.6

Take the square root of both side

v = √(1768.6)

v = 42.05 m/s

Therefore, the velocity with which the camera hits the ground is 42.05 m/s

3 0
3 years ago
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