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lana [24]
4 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]4 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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Explanation:

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A passenger walks from one side of a ferry to the other as it approaches a dock.Passenger's velocity is 1.55 m/s due north relat
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Answer:\theta =33.22^{\circ}

Explanation:

Given

Velocity of Passenger w.r.t to Ferry

v_{pf}=v_p-v_f=1.55\hat{j}-------1

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v_{pw}=v_p-v_w=4.5\left ( -\hat{i}+\hat{j}\right )--------2

Subtract 2 from 1

v_f-v_w=-4.5\hat{i}+2.95\hat{j}

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4 years ago
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Explanation:

6 0
3 years ago
Who proposed the laws of association to explain learning?
ohaa [14]
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I hope that helps my friend
4 0
4 years ago
A long, straight wire of radius R carries a steady current I that is uniformly distributed through the cross section of the wire
vitfil [10]

Answer:

a

  When r \ge R

      B =  \frac{ \mu_o *  I}{ 2 \pi r }

b

 When r< R

   B =  [\frac{\mu_o *  I }{ 2 \pi R^2} ]* r

Explanation:

From the question we are told that

   The  radius is  R  

   The  current is  I

    The  distance from the center

Ampere's law is mathematically represented as

       B[2 \pi r]  =  \mu_o  *  \frac{I r^2  }{R^2 }

      B =  \frac{ \mu_o}{2 \pi }  *  \frac{r}{R^2}

When r \ge R

=>     B =  \frac{ \mu_o *  I}{ 2 \pi r }

But when r< R

   B =  [\frac{\mu_o *  I }{ 2 \pi R^2} ]* r

     

4 0
4 years ago
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