Un péndulo de 4 m de longitud tiene una frecuencia de 5Hz. Calcular la longitud de otro péndulo que en el mismo lugar tiene una frecuencia de 4Hz

Answers

Answer 1

Conociendo la longitud y frecuencia de un péndulo, queremos encontrar la longitud de otro pendulo de tal forma que tenga otra frecuencia.

Veremos que la longitud del nuevo péndulo debe ser 6.25m

Sabemos que un péndulo de 4m de longitud tiene una frecuencia de 5Hz.

La frecuencia de un péndulo está dada por:

[tex]f = \frac{1}{2*\pi} *\frac{g}{l}[/tex]

Donde g es la aceleración gravitatoria y l es la longitud del péndulo, remplazando los datos que tenemos en esa ecuación obtenemos:

[tex]5 Hz = \frac{1}{2*3.14} *\sqrt{\frac{g}{4m} } \\\\(5Hz*2*3.14)^2*4m = g = 3,943.8 m/s^2[/tex]

Ahora debemos encontra la longitud de tal forma que la frecuencia sea 4Hz, entonces debemos resolver:

[tex]4Hz = \frac{1}{2*3.14} *\sqrt{ \frac{3943.8m/s^2}{l} }\\\\4hz*2*3.14 = \sqrt{ \frac{3943.8m/s^2}{l} }\\\\l =\frac{3943.8m/s^2}{ (4hz*2*3.14)^2} = 6.25m[/tex]

La longitud del nuevo péndulo deve ser 6.25m

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Answers

Answer:

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Explanation:

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Therefore, the velocity vs. time equation of the object is [tex]\displaystyle v=-\sin\left(\frac{2\pi}{3}t\right)\cdot \frac{2\pi}{3}[/tex].

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Answer:

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Answer:

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Please HELP!!

Diagram's BELOW↓

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b. 10 N up
c. 10 N down
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Answers

1. 20
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Answers

We have that for the Question, it can be said that With respect to this axis, the magnitude of the torque due to the weight and ,the thrust is

TW=19740N-mTT=130387.39N-m

From the question we are told

The figure shows a jet engine suspended beneath the wing of an airplane. The weight of the engine is 14900 N and acts as shown in the figure. In flight the engine produces a thrust of 61500 N that is parallel to the ground. The rotational axis in the figure is perpendicular to the plane of the paper. With respect to this axis, find the magnitude of the torque due to (a) the weight and (b) the thrust.

a)

Generally the equation for the Torque due to weight  is mathematically given as

[tex]TW=Engine weight*2.50*sin32\\\\TW=14900*2.50*sin32[/tex]

TW=19740N-m

b)

Generally the equation for the Torque due to thrust  is mathematically given as

[tex]TT=Engine thrust*2.50*cos32\\\\TT=61500*2.50*cos32\\\\[/tex]

TT=130387.39N-m

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We have that for the Question "Light of 650 nm wavelength illuminates two slits that are 0.20 mm  apart. (Figure 1) shows the intensity pattern seen on a screen behind the slits.

What is the distance to the screen?" it can be said that  the distance to the screen

d=1.168m

From the question we are told

Light of 650 nm wavelength illuminates two slits that are 0.20 mm

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What is the distance to the screen?

Generally the equation for the distance  is mathematically given as

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Answers

(1) As the angle of the ramp is increased, the normal force decreases.

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Let the angle of inclination of the ramp = θ

(1)

The normal force on an object on the ramp inclined to the ramp is calculated as follows;

[tex]F_n = mgcos (\theta)[/tex]

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[tex]F_n = mgcos (0)\\\\F_n = mg[/tex]

when θ is 90;

[tex]F_n = mgcos(90)\\\\F_n = 0[/tex]

Thus, as the angle of the ramp is increased, the normal force decreases.

(2)

The parallel force on an object on the ramp inclined to the ramp is calculated as follows;

[tex]F_x = mgsin(\theta)\\\\[/tex]

when θ is 0;

[tex]F_x = mgsin(\theta)\\\\F_x = mgsin(0) \\\\F_x = 0[/tex]

when θ is 90;

[tex]F_x = mgsin(90)\\\\F_x = mg[/tex]

Thus, as the angle of the ramp is increased, the parallel force increases.

(3)

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[tex]F_n = \mu F_n[/tex]

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(4)

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Answers

Answer:

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v²=15²-2*9.8*2.06

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v²=184.624

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v=13.59m/s

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Answers

By Newton's second law, the net force on the crate is

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300 N - f = (45 kg) (4.44 m/s²)

f = 300 N - (45 kg) (4.44 m/s²)

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the temperature remains constant because the given heat is used to increase only the potential energy of the molecules of the given substance (which results in the change of its state) and not the kinetic energy as temperature of a body is actually the measure of the average kinetic energy of its molecules.
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