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Maksim231197 [3]
3 years ago
5

WILL MARK BRAINLIESTFor the simple harmonic motion equation d =4sin(8pit) what is the period?

Physics
1 answer:
shtirl [24]3 years ago
4 0
The period of any wave is the time it takes for its angle
to go from zero to 2pi .

The 'sin' function is a wave.  The angle of this one is (8pi t).

When t=0, the angle is zero.
Wonderful.
Now, how long does it take for the angle to grow to 2pi ?

I*n other words, when is (8pi t) = 2pi ?

Divide each side by '2pi': . . . . . 4 t = 1

Divide each side by ' 4 ': . . . . . t = 1/4

And there you are.  Every time 't' grows by 1/4, (8pi t) grows by 2pi.
So if you graph this simple harmonic motion described by 'd', you'll
see the graph wiggle up and down with a period of 1/4 . 
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3 years ago
A 25 kg child plays on a swing having support ropes that are 2.20 m long. A friend pulls her back until the ropes are 42◦ from t
Semmy [17]

Answer:

A) P.E = 138.44 J

B) The velocity of swing at bottom, v = 3.33 m/s

C) The work done, W = -138.44 J

Explanation:

Given,

The mass of the child, m = 25 Kg

The length of the swing rope, L = 2.2 m

The angle of the swing to the vertical position, ∅ = 42°

A) The potential energy at the initial position ∅ = 42° is given by the relation

                                P.E = mgh joule

Considering h  = 0 for the vertical position

The h at ∅ = 42° is  h = L (1 - cos∅)

                               P.E = mgL (1 - cos∅)

Substituting the given values in the above equation

                               P.E = 25 x 9.8 x 2.2 (1 - cos42°)

                                      = 138.44 J

The potential energy for the child just as she is released, compared to the potential energy at the bottom of the swing is, P.E = 138.44 J

B) The velocity of the swing at the bottom.

At bottom of the swing the P.E is completely transformed into the K.E

                  ∴                 K.E = P.E

                                     1/2 mv² = 138.44

                                     1/2 x 25 x v² 138.44

                                            v² = 11.0752

                                             v = 3.33 m/s

The velocity of the swing at the bottom is, v = 3.33 m/s

C) The work done by the tension in the rope from initial position to the bottom

             Tension on string, T = Force acting on the swing, F

                      W=L\int\limits^0_\phi{F} \, d \phi

                             =L\int\limits^0_\phi{mg.sin \phi} \, d \phi

                            = -Lmg[cos\phi]_{42}^{0}

                            = - 2.2 x 25 x 9.8 [cos0 - cos 42°]

                            = - 138.44 J

The negative sign in the in energy is that the work done is towards the gravitational force of attraction.

The work done by the tension in the ropes as the child swings from the initial position to the bottom of the swing, W = - 138.44 J

3 0
3 years ago
When u write on a piece of glass sheet with a piece of chalk , the writing is not clear explain .
svetlana [45]
Becuse your weighting with chalk that has pigment
6 0
3 years ago
Describe the evolution of a pulsar over time, in particular how the rotation and pulse signal changes over time.
madam [21]

Answer:

As beams of particles and their associated energy are given off, the pulsar will lose energy slowly, which will decrease the rate of its rotation. The frequency of pulses would therefore decrease, so that fewer pulses are observed in a given time span. The strength of the pulse signal will also decrease so the pulses will become fainter. Eventually, the pulsar should rotate so slowly and have such a low emission of radiation that it would no longer be observable.

3 0
3 years ago
An object of mass 8kg moved around the circle of radius 4m. with a constant speed of 15m/s.
Marianna [84]

Answer:

a. Angular velocity = 0.267rad/s.

b. Centripetal acceleration = 56.25m/s.

Explanation:

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

Mass, m = 8kg

Radius, r = 4m

Constant speed, V = 15m/s

a. To find the angular velocity

Angular velocity = radius/speed

Substituting into the equation, we have;

Angular velocity = 4/15

Angular velocity = 0.267rad/s

b. To find the acceleration;

Centripetal acceleration = V²/r

Substituting into the equation, we have;

Centripetal acceleration = 15²/4

Centripetal acceleration = 225/4

Centripetal acceleration = 56.25m/s.

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3 years ago
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