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Anna35 [415]
4 years ago
14

a) ¿En qué posición es mínima la magnitud de la fuerza sobre la masa de un sistema masa-resorte? 1) x 0, 2) x A o 3) x A. ¿Por q

ué? b) Con m 0.500 kg, k 150 N/m y A 0.150 m, calcule la magnitud de la fuerza sobre la masa y la aceleración de la masa en x 0, 0.050 m y 0.150 m.
Physics
1 answer:
Tatiana [17]4 years ago
7 0

Answer:

a) the correct answer is 1 , b) x=0   F=0, a=0

x= 0.050    F= -7.5 N,  a= -15 m/s²

x= 0.150     F= 22.5 N,  a=- 45 m/s²

Explanation:

a) In a mass - spring system the force is given by the Hooke force,

          F = - k x

Analyzing this equation we see that the outside is proportional to the elongation from the equilibrium position, therefore the force is zero when the spring is in its equilibrium position

the correct answer is 1

b) we assume that the given values ​​are from the equilibrium position of the spring.

Let's calculate the force

x = 0

      F = 0

x = 0.050

      F = - 150 0.050

      F = - 7.50 N

x = 0.150

      F = - 150 0.150

      F = - 22.5 N

let's use Newton's second law to find the acceleration

      F = m a

      a = F / m

x = 0 m

      a = 0

x = 0.050 m

      a = -7.50 / 0.50

      a = - 15 m / s²

x = 0.150 m

      a = - 22.5 / 0.50

      a = - 45 m/s²

TRASLATE

a) En un sistema masa – resorte  la fuerza es dada por la fuerza de Hoke,  

          F= - k x

analizando esta ecuación vemos que la fuera es proporcional a la elongación desde la posición de equilibrio, por lo tanto la fuerza es cero cuando el resorte esta en su posición de equilibrio

la respuesta correcta es  1

b)suponemos que los valores dados son desde la posición de equilibrio del resorte.

Calculemos la fuerza  

x=0  

              F= 0

x=0.050  

              F = - 150 0.050

              F= - 7.50 N

x= 0.150  

                F= - 150 0.150

                F= - 22.5 N

usemos la segunda ley de Newton para encontrar la aceleración

          F = m a

          a = F/m

x =0  m

        a = 0

x= 0.050 m

         a = -7.50/ 0.50

          a =- 15 m/s²

x= 0.150 m

          a= - 22.5 / 0.50

          a= - 45 m/s²

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bulgar [2K]

Answer:

312,497.5Joules

Explanation:

Work done = force × distance

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F = 1250×9.8

F = 12250N

Get the distances using the equation of motion

v² = u² +2gS

30² =20²+2(9.8)S

900 =400+19.6S

900-400 =19.6S

500 = 19.6S

S = 500/19.6

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Work done = 12250×25.51

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6 0
3 years ago
Vision is blurred if the head is vibrated at 29 Hz because the vibrations are resonant with the natural frequency of the eyeball
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The effective spring constant of the system is 39.6 N/m

Explanation:

The frequency of oscillation of a spring-mass system is given by

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

where

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m is the mass

In this problem, we have:

f = 29 Hz is the frequency of vibration of the eyeball system

m = 7.5 g = 0.0075 kg is the mass

We can therefore re-arrange the equation to find the effective spring constant of the system. We find:

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Tarzan is testing the strength of a particular vine, which is 7 m long. As he is hanging on the vine, what is the magnitude of t
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Answer:

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Explanation:

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given,

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3 years ago
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sdas [7]
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______________________________________________________
Explanation:
______________________________________________

(2 * 10⁴ cm) * (4 * 10⁻¹² cm)  =

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