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 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?

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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

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H= (225*0.179)/19.6

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To find the velocity at maximum height:

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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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Light of 650 nm wavelength illuminates two slits that are 0.20 mm
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PLEAsE URGENT. thank you

Answers

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

Therefore

The distance to the screen

d=1.168m

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A 0.750 kg hammer is moving horizontally at 7.00 m/s when it strikes a nail and comes to rest after driving it 1.00 cm into a board. (a) Calculate the duration of the impact in seconds. (Enter a number.) s (b) What was the average force in newtons exerted downward on the nail

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Which of the following statements about migration is true?

a. Migration is always from one region to another.
b. Animals always migrate within a region.
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Answer:

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1. As the angle of the ramp is increased, the normal force increases /decreases / remains the same and the friction-force increases /decreases / remains the same. [1 Point] 2. As the angle of the ramp is increased, the force parallel increases /decreases / remains the same. [1 Point] 3. The angle at which the force down the plane was equal to the force of friction (for the cabinet) was _____________. [1 Point] 4. Consider a very low (~ zero) friction, 5.0 kg skateboard on a ramp at an angle of 15o to the horizontal. What would be the net force that would cause acceleration when the skateboard is allowed to move

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]

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.

(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)

The force of friction is calculated as follows;

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

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

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

Thus, the angle is zero degree

(4)

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[tex]F_k = Fn\\\\F_k = mg cos(\theta)\\\\F_k = 5 \times 9.8 \times cos(15)\\\\F_k = 47.33 \ N[/tex]

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A football is kicked with a velocity of 5 m/s at an angle of 53° above the horizontal. What is its speed at the
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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
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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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Explanation:

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Answer:

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hopefully this'll help

have a nice day!!! :D

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Answer:

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