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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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Student 1 lifts a box with a force of 500 N and sets it on a tabletop 1.2 m high. Student 2 pushes an identical box up a 5 m ram
Troyanec [42]

The student who did the most work is student 2 with 2500 Joules.

<u>Given the following data:</u>

  • Force 1 = 500 Newton
  • Distance 1 = 1.2 meter
  • Force 2 = 500 Newton
  • Distance 2 = 5 meter

To determine which of the students did the most work:

Mathematically, the work done by an object is given by the formula;

Work\;done = Force \times distance

<u>For </u><u>student 1</u><u>:</u>

Work\;done = 500 \times 1.2

Work done = 600 Joules

<u>For </u><u>student 2</u><u>:</u>

Work\;done = 500 \times 5

Work done = 2500 Joules.

Therefore, the student who did the most work is student 2 with 2500 Joules.

Read more: Read more: brainly.com/question/13818347

7 0
3 years ago
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Variations in the resistivity of blood can give valuable clues about changes in various properties of the blood. Suppose a medic
Andre45 [30]

Answer:

ρ = 1.70 Ωm

Explanation:

given data

diameter = 1.8 mm

distance = 4.8 cm

volt = 9 V

potential difference = 280

solution

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9 = 280 ×10^{-6} × \frac{\rho \times 0.048}{\pi \times 0.9^2 \times 10^{-6} }  

ρ = 1.70 Ωm

4 0
3 years ago
calculate the acceleration of a 41-kg crate of baseball gear when pulled sideways with a net force of 1255 N. neglect frictional
lara31 [8.8K]

In his celebrated Second Law of Motion, Newton wrote:

             Net force  =  (mass) x (acceleration).

By the process of "plugging in numbers", we can write

             1255 N  =  (41 kg) x (acceleration)

Now, after dividing each side by (41 kg), we have

             (1255 N) / (41 kg)  =  acceleration.

But  (1255N)/(41kg) = 30.61 m/s² .

So unless we have carelessly blooped the calculations somewhere,
that  30.61 m/s²  is the answer we're looking for.
  
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Slav-nsk [51]
<h3>An electric current consists of moving</h3>

<h2>ANSWER</h2>

A) atoms

<h3><em><u>B) electrons</u></em></h3>

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