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Brums [2.3K]
4 years ago
12

A block is on a frictionless table, on earth. The block accelerates at 7.5 m/s when a 70 N horizontal force is applied to it. Th

e block and table are set up on the moon. The acceleration due to gravity at the surface of the moon is 1.62 m/s. The weight of the block on the moon is closest to: O 9.5 N O 13 N O 11 N O 15 N O 7.7 N
Physics
1 answer:
Liono4ka [1.6K]4 years ago
3 0

Answer:

The weight of the block on the moon is 15 kg.

Explanation:

It is given that,

The acceleration of the block, a = 7.5 m/s²

Force applied to the box, F = 70 N

The mass of the block will be, m=\dfrac{F}{a}

m=\dfrac{70\ N}{7.5\ m/s^2}

m = 9.34 kg

The block and table are set up on the moon. The acceleration due to gravity at the surface of the moon is 1.62 m/s². The mass of the object remains the same. It weight W is given by :

W=m\times g

W=9.34\ kg\times 1.62\ m/s^2

W = 15.13 N

or

W = 15 N

So, the weight of the block on the moon is 15 kg. Hence, this is the required solution.

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Answer:

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.

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(a) -1.48\cdot 10^{-5}N

The electrostatic force exerted between the two sphere is given by:

F=k\frac{q_1 q_2}{r^2}

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r is the separation between the centres of the two spheres

In this problem,

q_1 = 12\cdot 10^{-9} C\\q_2 = -23\cdot 10^{-9} C\\r = 0.41 m

Substituting these values into the equation, we find the force

F=(9\cdot 10^9 Nm^2 C^{-2} )\frac{(12\cdot 10^{-9}C)(-23\cdot 10^{-9} C)}{(0.41 m)^2}=-1.48\cdot 10^{-5}N

And the negative sign means the force is attractive, since the two spheres have charges of opposite sign.

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When the two spheres are connected, the charge distribute equally over the two spheres (since they are identical, they have same capacitance), so each sphere will have a charge of

q=\frac{Q}{2}=\frac{-11\cdot 10^{-9}C}{2}=-5.5\cdot 10^{-9}C

So the electrostatic force between the two spheres will now be

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F=(9\cdot 10^9 Nm^2 C^{-2} )\frac{(-5.5\cdot 10^{-9} C)^2}{(0.41 m)^2}=+1.62\cdot 10^{-6}N

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