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1 year ago

Which of these periodic motions are simple harmonic?

1 answer:

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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A man of mass 85 kg runs up a flight of stairs of height 4.6 m in a time period

seraphim [82]

Explanation:

a) Power = work / time = force × distance / time

P = Fd/t

P = (85 kg × 9.8 m/s²) (4.6 m) / (12 s)

P ≈ 319 W

b) P = Fd/t

0.70 (319 W) = (m × 9.8 m/s²) (4.6 m) / (9.6 s)

m = 47.6 kg

7 0

2 years ago

At which point does the planet have the least gravitational force acting on it?

Elza [17]

Answer:

At which point does the planet have the least gravitational force acting on it?

Explanation:

In an elliptical orbit, when a planet is at its furthest point from the Sun, it is under the least amount of gravity, meaning that the force of gravity is strongest when it is closest.

5 0

2 years ago

Read 2 more answers

A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. its speed is 40 m/s when i

Alenkinab [10]

Given that,

Initial velocity , Vi = 0

Final velocity , Vf = 40 m/s

Acceleration due to gravity , a = 9.81 m/s²

Distance can be calculated as,

2as = Vf² - Vi²

2 * 9.81 *s = 40² - 0²

s = 81.55 m

For half height, that is, s = 40.77m

Vf= ??

2as = Vf² - Vi²

2 * 9.81 * 40.77 = Vf² - 0²

Vf² = 800

Vf = 28.28 m/s

Therefore, speed of roller coaster when height is half of its starting point will be 28 m/s.

5 0

2 years ago

A 1200.0-kg car is traveling at 19m/s. The driver suddenly slams on the brakes and skids to a stop. The coefficient of kinetic f

Alja [10]

<h2>Answer</h2>

option D)

2.4 seconds

<h2>Explanation</h2>

Given in the question,

mass of car = 1200kg

speed of car = 19m/s

Force due to direction of travel

F = ma

= 12000(a)

Force to due frictional force in reverse direction

-F = mg(friction coefficient)

= -12000(9.81)(0.8)

<h2>-mg(friction coefficient) = ma </h2>

(cancelling mass from both side of equation)

g(0.8) = a

(9.81)(0.8) = a

a = 7.848 m/s²

<h2>Use Newton Law of motion</h2><h3>vf - vo = a • t</h3>

where vf = final velocity

vo = initial velocity

a = acceleration

t = time

0 - 19 = 7.8(t)

t = 19/7.8

= 2.436 s

≈ 2.4s

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2 years ago

A charged object traveling 7 m in a uniform electric field of 5 N/C experiences a 4 J increase in Kinetic Energy.

Travka [436]

To solve this problem it is necessary to apply the principles of conservation of Energy in order to obtain the final work done.

The electric field in terms of the Force can be expressed as

$E = \frac{F}{q} \rightarrow F=Eq$

Where,

F = Force

E= Electric Field

q = Charge

Puesto que el trabajo realizado es equivalente al cambio en la energía cinetica entonces tenemos que

KE = W

KE = F*d

In the First Case,

$4 = (qE)d\\q = \frac{4}{Ed}\\q = \frac{4}{5*7}\\q = 0.1142C$

In Second Case,

$KE = q E'd$

$KE = (0.1142)(40)(7)$

$KE = 31.976J$

The total energy change would be subject to,

$\Delta KE = 31.976-4$

$\Delta KE = 27.976J$

Therefore the Kinetic Energy change of the charged object is 27.976J

3 0

2 years ago