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Fudgin [204]
3 years ago
7

Hold one end of a meter stick down firmly on a table so that 20 centimeters of the meter stick extends past the edge of the tabl

e. Pluck the end of the meter stick that extends past the table to produce a vibration and a sound as shown. Observe the vibration and sound of the meter stick. PLEASE HELP !
Physics
1 answer:
Usimov [2.4K]3 years ago
7 0
Just describe a low-sounding vibration. Give it some originality, something that sounds plausible. We can't really help you with this question because it's a personal point of view.
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Un objeto que tiene como masa 80 kg acelera a razón de 10 m/s2 ¿que fuerza lo impulzo
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Establish the relationship between acceleration and force ​
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Answer:

Newton's second law of motion describes the relationship between force and acceleration. They are directly proportional. If you increase the force applied to an object, the acceleration of that object increases by the same factor. In short, force equals mass times acceleration.

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A circuit contains two devices that are connected in parallel. If the resistance of one of these devices is 12 ohms and the resi
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<u>Answer</u>

3 Ohms

<u>Explanation</u>

when the resistors are in series, the resistance in the circuit increases. For example, if two resistors, R1 and R2 are in series, the combined resistance is R1+R2.

When connected in parallel, the total resistance is the reciprocal of (1/R1 + 1/R2)

In this case the resistors are in parallel.

Total resistance = (1/12 + 1/4)⁻¹

= (1/3)⁻¹

= 3 Ohms

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Which of the following nuclear configurations would be most stable?
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3 years ago
A 1.2 kg block sliding on a horizontal frictionless surface is attached to a horizontal spring with k =480 N/m. Let x be the dis
Angelina_Jolie [31]

Answer:

(a). The frequency is 3.18 Hz.

(b). The amplitude of the block's motion is 0.255 m.

(c). The expression for x as a function of time is x=0.255\cos(19.9 t+\dfrac{\pi}{2})

Explanation:

Given that,

Mass of block = 1.2 kg

Spring constant = 480 N/m

Speed = 5.2 m/s

We need to calculate the frequency

Using formula of frequency

f=\dfrac{1}{2\pi}\sqrt{\dfrac{480}{1.2}}

f=3.18\ Hz

The frequency is 3.18 Hz.

(b). We need to calculate the amplitude of the block's motion

Using relation of equation of amplitude and kinetic energy

\dfrac{1}{2}\times kA^2=\dfrac{1}{2}\times mv^2

Put the value into the formula

\dfrac{1}{2}\times500\times A^2=\dfrac{1}{2}\times1.2\times(5.2)^2

A^2=\dfrac{1.2\times(5.2)^2}{500}

A=\sqrt{\dfrac{1.2\times(5.2)^2}{500}}

A=0.255\ m

The amplitude of the block's motion is 0.255 m.

(c). We need to write the expression for x as a function of time

x=A\cos(\omega t+\phi)

Put the value into the equation

x=0.255\cos(19.9 t+\dfrac{\pi}{2})

The expression for x as a function of time is x=0.255\cos(19.9 t+\dfrac{\pi}{2})

Hence, This is the required solution.

6 0
3 years ago
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