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Ludmilka [50]
2 years ago
6

A roller coaster of mass 2000 kg is rolling down a track with an instantaneous speed 10m/s what is kinetic energy in Joules

Physics
1 answer:
Ira Lisetskai [31]2 years ago
3 0

The kinetic energy is 100,000 Joules, if a roller coaster of mass 2000 kg is rolling down a track with an instantaneous speed 10 m/s.

Explanation:

The given is,

              Mass - 2000 kg

              Instantaneous speed - 10 meter per second

Step: 1

            Formula to calculate the kinetic energy,

                          K.E = \frac{1}{2} mv^{2}....................(1)

           Where, K.E - Kinetic energy in joules

                           m - Mass in kg

                            v - Velocity or speed in meter per second

Step: 2

          From the give,

                  m = 2000 kg

                   v = 10 m/s

         Equation (1) becomes,

              K.E = \frac{1}{2} (2000)(10)^{2}

                       = (0.5)(2000)(100)

                       = 100000

                 K.E = 100,000 Joules

Result:

      The kinetic energy is 100,000 Joules, if a roller coaster of mass 2000 kg is rolling down a track with an instantaneous speed 10 m/s.

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v = 400m/58s = 6.9m/s

v = 400m/60s = 6.7m/s

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2 years ago
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Un lector de DVD, la velocidad de giro es de 5400 rpm. determina el valor velocidad angular en rad/s,la frecuencia y el periodo
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Responder:

A) ω = 565.56 rad / seg

B) f = 90Hz

C) 0.011111s

Explicación:

Dado que:

Velocidad = 5400 rpm (revolución por minuto)

La velocidad angular (ω) = 2πf

Donde f = frecuencia

ω = 5400 rev / minuto

1 minuto = 60 segundos

2πrad = I revolución

Por lo tanto,

ω = 5400 * (rev / min) * (1 min / 60s) * (2πrad / 1 rev)

ω = (5400 * 2πrad) / 60 s

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Dado que :

ω = 2πf

donde f = frecuencia, ω = velocidad angular en rad / s

f = ω / 2π

f = 565.56 / 2π

f = 90.011669

f = 90 Hz

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Recordar T = 1 / f

Por lo tanto,

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2 years ago
A nonuniform, 80.0-g, meterstick balances when the support is placed at the 51.0-cm mark. At what location on the meterstick sho
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Explanation:

Given

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balancing torque

80\times 10^{-3}(51-50)=5\times 10^{-3}(50-x)

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x=\frac{170}{5}

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