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Natali [406]
2 years ago
12

(1 point)

Physics
1 answer:
jarptica [38.1K]2 years ago
4 0
Sorry, I can’t access the picture
You might be interested in
A metal cylinder with a mass of 4.20 kg is attached to a spring and is able to oscillate horizontally with negligible friction.
kherson [118]

Answer:

a) k = 120 N / m

, b)    f = 0.851 Hz

, c)  v = 1,069 m / s

, d)  x = 0

, e)  a = 5.71 m / s²

, f)   x = 0.200 m

, g)  Em = 2.4 J

, h) v = -1.01 m / s

Explanation:

a) Hooke's law is

         F = k x

         k = F / x

          k = 24.0 / 0.200

          k = 120 N / m

b) the angular velocity of the simple harmonic movement is

        w = √ k / m

        w = √ (120 / 4.2)

        w = 5,345 rad / s

Angular velocity and frequency are related.

       w = 2π f

        f = w / 2π

        f = 5.345 / 2π

        f = 0.851 Hz

c) the equation that describes the movement is

        x = A cos (wt + Ф)

As the body is released without initial velocity, Ф = 0

        x = 0.2 cos wt

Speed ​​is

       v = dx / dt

       v = -A w sin wt

The speed is maximum for sin wt = ±1

       v = A w

       v = 0.200 5.345

       v = 1,069 m / s

d) when the function sin wt = -1 the function cos wt = 0, whereby the position for maximum speed is

       x = A cos wt = 0

       x = 0

e) the acceleration is

       a = d²x / dt² = dv / dt

       a = - Aw² cos wt

The acceleration is maximum when cos wt = ± 1

       a = A w²

        a = 0.2   5.345

        a = 5.71 m / s²

f) the position for this acceleration is

       x = A cos wt

       x = A

       x = 0.200 m

g) Mechanical energy is

        Em = ½ k A²

        Em = ½ 120 0.2²

       Em = 2.4 J

h) the position is

         x = 1/3 A

Let's calculate the time to reach this point

         x = A cos wt

        1/3 A = A cos 5.345t

         t = 1 / w cos⁻¹(1/3)

The angles are in radians

t = 1.23 / 5,345

t = 0.2301 s

Speed ​​is

v = -A w sin wt

v = -0.2 5.345 sin (5.345 0.2301)

v = -1.01 m / s

i) acceleration

a = -A w² sin wt

a = - 0.2 5.345² cos (5.345 0.2301)

      a = -1.91 m / s²

5 0
3 years ago
A mass of 150 g stretches a spring 1.568 cm. If the mass is set in motion from its equilibrium position with a downward velocity
nadezda [96]

Answer:

u(t)=\frac{1}{5} sin\ (25t)

Explanation:

Given:

  • mass of the body stretching the spring, m=150\ g
  • extension in spring, \Delta x=1.568\ cm
  • velocity of oscillation, u'(0)=20\ cm.s^{-1}
  • initial displacement position of equilibrium, u(0)=0

<u>According to given:</u>

m.g=k.\Delta x

150\times 980=k\times 1.568

k=93750\ dyne.cm^{-1}

<u>we know frequency:</u>

\omega=\sqrt{\frac{k}{m} }

\omega=\sqrt{\frac{93750}{150} }

\omega=25

Now, for position of mass in oscillation:

u= A.sin\ (\omega.t)+B.cos\ (\omega.t)

u= A.sin\ (25.t)+B.cos\ (25.t)

at t=0;\ u(0)=0\ \Rightarrow A=0

∴u(t)=B.sin\ (25.t)

∵ at t=0;\ u'(0)=20\ cm.s^{-1}\ \Rightarrow B=\frac{1}{5}

u(t)=\frac{1}{5} sin\ (25t)

7 0
3 years ago
"Suppose you tie a rock to the end of a 0.96 m long string and spin it in a horizontal circle with a constant angular velocity o
Ber [7]

Answer:

The mass of the rock is  m = 2.46406 \ kg

Explanation:

From the question we are told that

   The length of the string is  l  = 0.96 \ m

    The angular velocity is  w =  20.25 \ rad/s

     The tension on the string is T  = 970 \ N

Generally the centripetal force acting on the rock is mathematically evaluated as

       T = mlw^2

making m the subject of the formula

      m = \frac{T}{w^2 * l}

substituting values

     m = \frac{970}{(20.25^2) * (0.96)}

     m = 2.46406 \ kg

       

8 0
3 years ago
34.9x46x809 Please helpp
Fed [463]
<h2>34.9×46×809</h2><h3>=1605.4×809 </h3><h3>=1298768.6</h3>

please mark this answer as brainlist

6 0
3 years ago
What would be the speed of an object just before hitting the ground if dropped 100 meters?
PolarNik [594]

<u>We are given:</u>

the initial height of the object (h) = 100 m

initial velocity (u) = 0 m/s

we will let the value of g = 10 m/s/s

<u>Speed of the object just before hitting the ground:</u>

From the third equation of motion:

v² - u² = 2ah     (where v is the final velocity)

replacing the variables, we get:

v² - (0)² = 2(10)(100)

v² = 2000

v = 10√20 = 44.7 m/s

Therefore, the speed of the object just before hitting the ground is 44.7 m/s

8 0
3 years ago
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