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Nina [5.8K]
3 years ago
10

What is the magnitude of the force excepted by the mass on the spring scale?

Physics
1 answer:
pentagon [3]3 years ago
4 0

Answer: Option c

Explanation:

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In a region of space where gravitational forces can be neglected, a sphere is accelerated by a uniform light beam of intensity 8
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The correct answer is B

Explanation:

To calculate the acceleration we must use Newton's second law

      F = m a

      a = F / m

To calculate the force we use the defined pressure and the radiation pressure for an absorbent surface

       P = I / c        absorbent surface

       P = F / A

       F / A = I / c

       F = I A / c

The area of ​​area of ​​a circle is

      A = π r²

We replace

     F = I π r² / c

Let's calculate

     F = 8.0 10⁻³ π (1.0 10⁻⁶)²/3 10⁸

     F = 8.375 10⁻²³ N

Density is

      ρ = m / V

      m = ρ V

      m = ρ (4/3 π r³)

      m = 4500 (4/3 π (1 10⁻⁶)³)

      m = 1,885 10⁻¹⁴ kg

Let's calculate the acceleration

     a = 8.375 10⁻²³ / 1.885 10⁻¹⁴

     a = 4.44 10⁻⁹ m/s²               absorbent surface

The correct answer is B

4 0
3 years ago
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
Tatiana [17]

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

7 0
3 years ago
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