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blsea [12.9K]
1 year ago
13

what is the magnitude of the gravitational force acting on the earth due to the sun? express your answer in newtons.

Physics
1 answer:
Alborosie1 year ago
5 0

The gravitational force the sun experiences from the earth is 3.48×10²²N, which is exactly the same as the force the sun experiences from the earth.

  • Gravity is a force that develops as a result of the attraction between mass-containing objects. The mass of the object has a direct relationship to the strength of this attraction. r equals the separation of two objects.

F = G (M₁M₂)/r²

Where, F  the gravitational force

G=6.67×10⁻¹¹Nm²kg⁻² gravitational constant

M₁=5.98×10²⁴kg  mass of earth

M₂= 1.99×10³⁰ kg the mass of the sun

r =15×10¹⁰ m is the distance between sun and earth

Putting all the values in above equation,

F = 6.67×10⁻¹¹Nm²kg⁻²(5.98×10²⁴kg 1.99×10³⁰ kg)/15×10¹⁰ m

On solving the above equation we get,

F = 3.48×10²²N

To know more about gravitational force

brainly.com/question/12830265

#SPJ4

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Can the validity of a model be limited, or must it be universally valid? How does this compare to the required validity of a the
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Describe an experiment to determine how the frequency of a vibrating string depends on the length of the string
Ksivusya [100]

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For a vibrating string, the fundamental frequency depends on the string's length, its tension, and its mass per unit length. ... The fundamental frequency of a vibrating string is inversely proportional to its length.

Explanation:

Sounds of a single pure frequency are produced only by tuning forks and electronic devices called oscillators; most sounds are a mixture of tones of different frequencies and amplitudes. The tones produced by musical instruments have one important characteristic in common: they are periodic, that is, the vibrations occur in repeating patterns. The oscilloscope trace of a trumpet's sound shows such a pattern. For most non-musical sounds, such as those of a bursting balloon or a person coughing, an oscilloscope trace would show a jagged, irregular pattern, indicating a jumble of frequencies and amplitudes.

A column of air, as that in a trumpet, and a piano string both have a fundamental frequency—the frequency at which they vibrate most readily when set in motion. For a vibrating column of air, that frequency is determined principally by the length of the column. (The trumpet's valves are used to change the effective length of the column.) For a vibrating string, the fundamental frequency depends on the string's length, its tension, and its mass per unit length.

In addition to its fundamental frequency, a string or vibrating column of air also produces overtones with frequencies that are whole-number multiples of the fundamental frequency. It is the number of overtones produced and their relative strength that gives a musical tone from a given source its distinctive quality, or timbre. The addition of further overtones would produce a complicated pattern, such as that of the oscilloscope trace of the trumpet's sound.

How the fundamental frequency of a vibrating string depends on the string's length, tension, and mass per unit length is described by three laws:

1. The fundamental frequency of a vibrating string is inversely proportional to its length.

Reducing the length of a vibrating string by one-half will double its frequency, raising the pitch by one octave, if the tension remains the same.

2. The fundamental frequency of a vibrating string is directly proportional to the square root of the tension.

Increasing the tension of a vibrating string raises the frequency; if the tension is made four times as great, the frequency is doubled, and the pitch is raised by one octave.

3. The fundamental frequency of a vibrating string is inversely proportional to the square root of the mass per unit length.

This means that of two strings of the same material and with the same length and tension, the thicker string has the lower fundamental frequency. If the mass per unit length of one string is four times that of the other, the thicker string has a fundamental frequency one-half that of the thinner string and produces a tone one octave lower.

7 0
3 years ago
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Explanation:

It is known that wave intensity is the power to area ratio.

Mathematically,    I = \frac{P}{A}

As it is given that power is 28.0 W and area is 7 \times 10^{-5} m^{2}.

Therefore, sound intensity will be calculated as follows.

             I = \frac{P}{A}

               = \frac{28.0 W}{4 \times 3.14 \times 7 \times 10^{-5} m^{2}}

                = 0.318 \times 10^{5} W/m^{2}

or,             = 3.18 \times 10^{4} W/m^{2}

Thus, we can conclude that sound intensity at the position of the microphone is 3.18 \times 10^{4} W/m^{2}.

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