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2 years ago

What is simple harmonic oscillation? express your answer verbally, mathematically, and visually.

1 answer:

3 0

Simple harmonic motion is the motion in objects where the restoring for d is directly proportional to the body displacement.

<h3>What is simple harmonic motion?</h3>

Simple harmonic motion a type of periodic motion where the force that restores moving object to it's right position is directly proportional to the magnitude of body's displacement and this normally acts towards the object's equilibrium position.

The formulate for simple harmonic motion are

,F = −kx, where F is the force, x is the displacement, and k is a constanst. An example of a simple harmonic oscillator is the vibration of a mass attached to a vertical spring.

Simple harmonic oscillator can be represented by the equation x ( t ) = A cos ⁡ ( 2 π f t ) x(t) = A\cos(2\pi f t) x(t)=Acos(2πft)x,

where t is period.

x is displacement

A is amplitude.

Learn more about simple harmonic motion here.

brainly.com/question/17315536

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The electric potential, when measured at a point equidistant from two particles that have charges equal in magnitude but of oppo

Reika [66]

Answer:

C) equal to zero

Explanation:

Electric potential is calculated by multiplying constant and charge, then dividing it by distance. The location that we want to measure is equidistant from two particles, mean that the distance from both particles is the same(r2=r1). The charges of the particle have equal strength of magnitude but the opposite sign(q2=-q1). The resultant will be:V = kq/r

ΔV= V1 + V2= kq1/r1 + kq2/r2

ΔV= V1 + V2= kq1/r1 + k(-q1)/(r)1

ΔV= kq1/r1 - kq1/r1

ΔV=0

The electric potential equal to zero

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4 years ago

Describe the role of observations in scientific endeavor and explain the characteristics of good observations.

koban [17]

Observations are used in order to collect data and record a variety of interesting or useful key points about the subject or specimen in observation. These observations, if made well, can be recorded and used to supplement a hypothesis.

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3 years ago

A car accelerates from rest at a constant rate of 3.5m/s^2. What is the velocity of the car 12s later?

Sladkaya [172]

I think 42… sorry if its wrong :( hope it helps..

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3 years ago

Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to the ground. Think about the amounts o

likoan [24]

Answer:

1) At the highest point of the building.

2) The same amount of energy.

3) The kinetic energy is the greatest.

4) Potential energy = 784.8[J]

5) True

Explanation:

Question 1

The moment when it has more potential energy is when the ball is at the highest point in the building, that is when the ball is at a height of 40 meters from the ground. It is taken as a point of reference of potential energy, the level of the soil, at this point of reference the potential energy is zero.

$E_{p} = m*g*h\\E_{p} = 2*9.81*40\\E_{p} = 784.8[J]$

Question 2)

The potential energy as the ball falls becomes kinetic energy, in order to be able to check this question we can calculate both energies with the input data.

$E_{p}=m*g*h\\ E_{p} = 2*9.81*20\\ E_{p} = 392.4[J]\\$

And the kinetic energy will be:

$E_{k}=0.5*m*v^{2}\\ where:\\v = velocity = 19.8[m/s]\\E_{k}=0.5*2*(19.8)^{2}\\ E_{k}=392.04[J]$

Therefore it is the ball has the same potential energy and kinetic energy as it is half way through its fall.

Question 3)

As the ball drops all potential energy is transformed into kinetic energy, therefore being close to the ground, the ball will have its maximum kinetic energy.

$E_{k}=E_{p}=m*g*h = 2*9.81*40\\ E_{k} = 784.8[J]\\ E_{k} = 0.5*2*(28)^{2}\\ E_{k} = 784 [J]$