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inna [77]
3 years ago
6

A block with a mass of 0.400 kg is connected to a spring, displaced in the positive direction a distance of 50.0 cm from equilib

rium, and released from rest at t = 0. The block then oscillates without friction on a horizontal surface. After being released, the first time the block is a distance of 25.0 cm from equilibrium is at t = 0.200 s.
(a) What is the block's period of oscillation?

(b) What is the the value of the spring constant?
Physics
1 answer:
Gemiola [76]3 years ago
7 0

Answer:

a)  T = 1,199 s b)  k = 11.0 N / m

Explanation:

The oscillatory movement of a mass-spring system is described by the equation

    x (t) = A cos (w t)

Where A is the amplitude of the movement. W is the angular velocity, which is given by

    w = √ k / m

The amplitude of the system is the maximum elongation that the spring has before releasing it, in this case it is A = 50 cm = 0.50 m, the equation is

     x = 0.50 cos (wt)

We can calculate the angular velocity with the point given by x = 0.250 m for t = 0.200s

     w t = cos⁻¹ (x / A)

     w = 1 / t cos⁻¹ (x / A)

     w = 1/0.200  cos⁻¹(0.25 / 0.50)

Let's be careful because the angle is in radians

     w = 5.24 rad / s

Angular velocity is

    w = 2 π f = 2π / T

    T = 2π / w

    T = 2π / 5.24

    T = 1,199 s

With the angular velocity equation we can take off the spring constant

    w² = k / m

    k = m w²

    k = 0.400 5.24²

    k = 11.0 N / m

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A) incandescent ligth bulb, its efficiency is about 10%

Explanation:

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Explanation:

For an isolated system as given the energy lost and gains in the system will be zero therefore sum of all transfer of energy will be zero,as the temperature will also remain same

A specific heat formula is given as                  

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For Copper:

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