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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A mover pushes a 45 kg crate across a level floor with a force of 300 N, but the crate accelerates at a rate of only 4.44 m/s2 because a friction force opposes the crate's motion. What is the magnitude of this force of friction? 300 N Force of Friction O A. 50 N N OB. 25 N C. 15 N 0 D. 100 N​

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

F = 300 N - f = (45 kg) (4.44 m/s²)

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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.

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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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A brick is thrown upward from the top of a building at an angle of 25 degrees to the horizontal with an initial speed of 15 m/s. If the brick is in flight for 4 s before it hits the ground, how tall is the building?

Answers

Answer:

First, find the maximum height, which according to the values given, can be stated as:

H=(u²sin²theta)/2g

u=15m/s, theta=25 degrees, g=9.8m/s²

H= (15² * (sin 25)²))/2*9.8

H= (225*0.179)/19.6

H= 40.275/19.6

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To find the velocity at maximum height:

Use the formula

v²=u²-2gH

It's minus because the brick was thrown upwards

So plugging everything into the above formula:

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v²=225–40.376

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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

apart. (Figure 1) shows the intensity pattern seen on a screen behind the slits.

What is the distance to the screen?

Generally the equation for the distance  is mathematically given as

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d=1.168m

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

(2) As the angle of the ramp is increased, the parallel force increases.

(3) The angle at which the force down the plane was equal to the force of friction is zero degree.

(4) The net force that would cause acceleration is 47.33 N.

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(1)

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

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

when θ is 0;

[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.

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

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

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[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]

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(3)

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[tex]F_k = \mu mgcos(\theta)[/tex]

[tex]F_k = \mu mg cos(0)\\\\F_k = \mu mg[/tex]

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Answers

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Got it right on Edge 2023. : )

Please HELP!!

Diagram's BELOW↓

1. What is the mass of the object experimented on in this situation?

a. 10 kg
b. 15 kg
c. 20 kg
d. 25 kg


2. What is the net force on this object?

a. 20 N to the left
b. 10 N up
c. 10 N down
d. 35 N to the right


Thank you in advance! Very appreciated! (I will mark brainliest) :)

Answers

1. 20
2. 35n to the right

The mass of the object experimented in the situation will be equal to 20 kg in each case. Hence, option C is correct.

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It is a vector quantity because it has both magnitude and direction.

The given data as per the question is,

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