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Ivanshal [37]
2 years ago
5

Which of these periodic motions are simple harmonic?

Physics
1 answer:
marta [7]2 years ago
4 0

Since the acceleration of the body is directed toward the center of the motion and proportional to its displacement from that point, option a and c are the correct answers of simple harmonic motion

SIMPLE HARMONIC MOTION

Simple Harmonic Motion is the periodic motion of particle or object along a straight line such that the acceleration of the body is directed toward the center of the motion and proportional to its displacement from that point.

Another name for periodic motion is oscillatory motion.

Example are

  • Motion of a simple pendulum.
  • Motion of mass suspended from spring.
  • Motion of loaded tube in a liquid.

A child swinging on a playground swing at angle Ø = 45° will experience back and forth movement which indeed is a simple harmonic motion.

A CD rotating in a player will experience a circular motion and not periodic motion

An oscillating clock pendulum (Ø = 10°) will also experience back and forth motion which is periodic motion and it is simple harmonic motion because it is directed toward a fixed point.

Therefore, option a and c are simple harmonic motion.

Learn more about Simple Harmonic Motion here: brainly.com/question/24646514

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Lubov Fominskaja [6]

question: Please help!!!

If a bottle is being squeezed with a force of 10 Newtons and the area of the bottle is 15

squared inches. What is the pressure??

Answer:

1025.64 N/m²

Explanation:

Pressure: This can be defined as the ratio of force to area. The si unit of pressure is N/m².

From the question,

P = F/A........................ Equation 1

Where F = Force, A = Area.

Given: F = 10 Newtons, A = 15 Squared Inches = (15×0.00065) = 0.00975 m²

Substitute these values into equation 1

P = 10/0.00975

P = 1025.64 N/m²

Hence the pressure of the bottle is 1025.64 N/m²

5 0
2 years ago
Read 2 more answers
Ron is on a Ferris wheel of radius 30 ft that turns counterclockwise at a rate of one revolution every 12 seconds. The lowest po
alexgriva [62]

Answer:

x = 30cos\frac{\pi}{6}t

y = 30sin\frac{\pi}{6}t + 45

Explanation:

1 full revolution is 2\pi. let \theta be the angle of Ron's position.

At t = 0. \theta = 0

one full revolution occurs in 12 sec, so his angle at t time is

\theta =2\pi \frac{t}{12} = \frac{\pi}{6}t

r is radius of circle and it is given as

x = rcos\theta

y = rsin\theta

for r = 30 sec

x = 30cos\frac{\pi}{6}t

y = 30sin\frac{\pi}{6}t

however, that is centered at (0,0) and the positioned at time t = 0 is (30,0). it is need to shift so that the start position is (30,45). it can be done by adding to y

x = 30cos\frac{\pi}{6}t

y = 30sin\frac{\pi}{6}t + 45

7 0
2 years ago
Can i eat air? im hungry
Lerok [7]

Answer:

Yes

Explanation:

8 0
2 years ago
Read 2 more answers
Through what process is carbon pulled from the atmosphere in the carbon cycle
Andrei [34K]

Carbon is pulled from the atmosphere in the carbon cycle through the process of photosynthesis. Details about photosynthesis can be found below.

<h3>What is photosynthesis?</h3>

Photosynthesis is the process whereby green plants obtain their nutrition by utilizing energy from sunlight.

Green plants absorb carbon in the form of carbon dioxide from the atmosphere and use it in the photosynthetic process.

This means that one way that carbon is removed from the atmosphere during the carbon cycle is through photosynthesis.

Learn more about photosynthesis at: brainly.com/question/1388366

#SPJ1

7 0
2 years ago
Two long straight parallel wires are separated by 7.0cm. There is a 2.0A current flowing in the first wire. If the magnetic fiel
aliina [53]

Answer:

The current in the second wire is 4.4 A.

Explanation:

Given that,

Distance =7.0 cm

Current in first wire = 2.0 A

The magnetic field strength is zero at distance of 2.2 cm from the first wire.

We need to calculate the current in the second wire

Using formula of magnetic field

B=B_{1}-B_{2}

0=\dfrac{\mu_{0}I_{1}}{2\pi r_{1}}-\dfrac{\mu_{0}I_{2}}{2\pi r_{2}}

\dfrac{\mu_{0}I_{1}}{2\pi r_{1}}=\dfrac{\mu_{0}I_{2}}{2\pi r_{2}}

\dfrac{I_{1}}{r_{1}}=\dfrac{I_{2}}{(d-r_{1})}

Here, r_{2}=d-r_{1}

I_{2}=\dfrac{I_{1}\times(d-r_{1})}{r_{1}}

Put the value into the formula

I_{2}=\dfrac{2.0\times(7.0-2.2)}{2.2}

I_{2}=4.4\ A

Hence, The current in the second wire is 4.4 A.

4 0
3 years ago
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