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vesna_86 [32]
2 years ago
9

There is a current of 0.99 a through a light bulb when its connected to a 9.7 v battery what is the resistance of the light bulb

Physics
1 answer:
denis23 [38]2 years ago
4 0

Ohm's law states that V=IR, where V=voltage, I=current(amps), and R=resistance (in Ohms).

Plugging the values into the above equation yields a resistance in the light bulb of 9.8 ohms

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Consider the rocket in the image. Which unbalanced force accounts for the direction of the net force of the rocket ?
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unbalenced force

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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
A 47-kg packing crate is pulled with constant speed across a rough floor with a rope that is at an angle of 37 degrees above the
levacccp [35]

Tension in the rope due to applied force will be given as

F = 142 N

angle of applied force with horizontal is 37 degree

displacement along the floor = 6.1 m

so here we can use the formula of work done

W = F d cos\theta

now we can plug in all values above

W = 142 * 6.1 * cos37

W = 691.8 J

So here work done to pull is given by 691.8 J


8 0
2 years ago
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