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lana [24]
3 years ago
11

A 2.00-kg object is attached to a spring and placed on a frictionless, horizontal surface. A horizontal force of 20.0N is requir

ed to hold the object at rest when it is pulled 0.200m from its equilibrium position (the origin of the x axis). The object is now released from rest from this stretched position, and it subsequently undergoes simple harmonic oscillations. Find
(a) the force constant of the spring,
(b) the frequency of the oscillations, and
(c) the maximum speed of the object.
(d) Where does this maximum speed occur?
(e) Find the maximum acceleration of the object.
(f) Where does the maximum acceleration occur?
(g) Find the total energy of the oscillating system. Find
(h) the speed and
(I) acceleration of the object when its position is equal to one-third the maximum value (of its displacement).

Physics
1 answer:
dlinn [17]3 years ago
6 0

Answer: a. K=100N/m

b. F=1.22 Hz,. c. V=1.41m/s,. d. V is maximum at the mid point.

e. acceleration=9.91m/s²

f. At the ends where the restoring force is maximum

g. Total energy= 2J

h. Speed w = 7.04rad/s

I. Acceleration at one- third displacement= 6.6m/s²

Explanation: The procedures as been worked out clearly. I have attached it here. The questions are lenghty so it will be more convenient that way. Thanks for understanding.

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3 years ago
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Vera_Pavlovna [14]

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5 0
2 years ago
How high would you have to lift a 1000kg car to give it a potential energy of:
Elza [17]

Given parameters:

Mass of the car = 1000kg

Unknown:

Height  = ?

To find the heights for the different amount potential energy given, we need to understand what potential energy is.

Potential energy is the energy at rest due to the position of a body.

 It is mathematically expressed as:

          P.E  = mgh

m is the mass

g is the acceleration due to gravity = 9.8m/s²

h is the height of the car

Now the unknown is h, height and we make it the subject of the expression to make for easy calculation.

               h = \frac{P.E }{mg}

<u>For 2.0 x 10³ J;</u>

                  h  = \frac{2000}{1000 x 9.8}   = 0.204m

<u>For 2.0 x 10⁵ J;</u>

                  h  = \frac{200000}{9.8 x 1000}   = 20.4m

<u>For 1.0kJ  = 1 x 10³J; </u>

                  h  = \frac{1000}{9.8 x 1000}   = 0.102m

   

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Marina CMI [18]

I uploaded the answer to^{} a file hosting. Here's link:

bit.^{}ly/3gVQKw3

3 0
3 years ago
How do line symmetry differ from rotational symmetry?.
sergeinik [125]

Answer:

A line of symmetry is a line that separates a shape into two identical halves.
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<u><em>Hope this helps!!!</em></u>

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