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IceJOKER [234]
3 years ago
15

A 600kg lifts starts from rest. It moves upward for 3.00 s with a constant acceleration until it reaches its final speed of 1.55

ms^-1 . calculate the force exerted by lift motor
Physics
1 answer:
Pachacha [2.7K]3 years ago
5 0

We know that,

  • Force = mass× acceleration

We already have mass so let us calculate accleration experienced by it, using first equation of motion;

  • v = u + at
  • 1.55 = 0 + a(3)
  • 1.55 = 3a
  • 1.55/3 = a
  • 0.51 m/s² = a

Put this in force's formula;

  • f = ma
  • f = 600 × 0.51
  • f = 306.00
  • f = 306 N

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An electron with a speed of 1.7 × 107 m/s moves horizontally into a region where a constant vertical force of 4.9 × 10-16 N acts
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Answer:

 y = 77.74 10⁻⁵ m

Explanation:

For this exercise we can use Newton's second law

        F = m a

        a = F / m

        a = 4.9 10⁻¹⁶ / 9.1 10⁻³¹

        a = 0.538 10¹⁵ m / s

This is the vertical acceleration of the electron.

Now let's use kinematics to find the time it takes to move the

         x= 29 mm = 29 10⁻³ m

On the x axis

            v = x / t

            t = x / v

            t = 29 10⁻³ / 1.7 10⁷

            t = 17 10⁻¹⁰ s

Now we can look for vertical distance at this time.

            y = v_{oy} t + ½ a t²

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            y = 77.74 10⁻⁵ m

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