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Cloud [144]
3 years ago
9

A bicycle tire is spinning counterclockwise at 2.70 rad/s. During a time period At 1.90 s, the tire is stopped and spun in the o

pposite (clockwise) direction also at 2.70 rad/s. Calculate the change in the tire's angular velocity Aw and the tire's average angular acceleration aav. (Indicate the direction with the signs of your answers.)
(a) the change in the tire's angular velocity Aco (in rad/s) rad/s
(b) the tire's average angular acceleration aav (in rad/s2) rad/s
Physics
1 answer:
Naddik [55]3 years ago
7 0

Answer:

a) -5.40 rad/s

b) -2.842 rad/s²

Explanation:

The direction is important in dealing with such questions. Clockwise is considered negative and counterclockwise is considered positive

a) Δω = final angular velocity - initial angular velocity

          = -2.70 rad/s - 2.70 rad/s

          = -5.40 rad/s

b) ∝ = Δω/Δt = (-5.40 rad/s)/1.90s = -2.842 rad/s²

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A person carries a mass of 10 kg and walks along the +x-axis for a distance of 100m with a constant velocity of 2 m/s. What is t
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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
4 years ago
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