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Zepler [3.9K]
2 years ago
5

SIMPLE HARMONIC MOTION 1.0 Objective To study simple harmonic motion by observing the motion of a simple pendulum. 2.0 Simple ha

rmonic motion Oscillatory motion is extremely common in nature. Examples include waves (water, sound, earthquake, etc.) and vibrations produced by musical instruments. If the oscillation is characterized by a constant frequency and amplitude (if the motion reproduces itself in a fixed time period T), then the motion is said to be "harmonic." If the oscillation can be described as a sinusoidal function of time and position, the motion is said to be "simple harmonic." Simple harmonic motion (SHM) occurs when for every applied force or torque, there is a restoring force or torque which is proportional to the displacement of the system from its equilibrium position. 2.0.1 Name two real-world examples of harmonic motion. (Hint: Southern California is known for what sort of natural disasters?)
Physics
1 answer:
bagirrra123 [75]2 years ago
8 0

Answer:

the waves in the sea,  leaves of the trees, cables in the bridges, pendulum clock

Explanation:

In nature there are many examples of simple harmonic motion, for example.

* The movement of the waves in the sea is an oscillation movement up and down

* The movement of the leaves of the trees when a wind blows and then stops, but the leaf and branches are oscillating

* The movement of the cables in the bridges, especially in the suspension bridges

* The movement of a pendulum clock

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enyata [817]

Answer:

See Below

Explanation:

Okay, I thinkkk what it is asking by what you summarzied for me issss:

They split the total time into four quarters. They then took (for the first quarter) the start time. Then when the first quarter ends and the second quarter starts is the "end" time.

They then subtract the start time of the second quarter from the end time of the first quarter.

I hope this helps, good luck! :D

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The velocity of the wave involved in the Doppler effect:
Margarita [4]

The Doppler Effect provides the equation for the calculation of apparent frequency:

f=fo[vo/(vo-vr)] 

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f=apparent frequency 
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The velocity of the doppler wave is 
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where λ is light wavelength. Hence,

v=λfo[vo/(vo-vr)] 

Based on the equation, we can say that wave velocity will always be defined by one and only one wavelength.

Therefore the answer is letter C.

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4 0
3 years ago
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Newton's first law of motion is the law of inertia. What does it state?
vladimir2022 [97]

Answer:

b is the right answer

Explanation:

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5 0
3 years ago
Air bags are designed to deploy in 10 ms. Given that the air bags expand 20 cm as they deploy, estimate the acceleration of the
joja [24]

As it is given that the air bag deploy in time

t = 10 ms = 0.010 s

total distance moved by the front face of the bag

d = 20 cm = 0.20 m

Now we will use kinematics to find the acceleration

d = v_i*t + \frac{1}{2}at^2

0.20 = 0 + \frac{1}{2}a*0.010^2

0.20 = 5 * 10^{-5}* a

a = 4000 m/s^2

now as we know that

g = 10 m/s^2

so we have

a = 400g

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3 0
3 years ago
A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch
Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: -3.62\cdot 10^7 N/C

c) Electric field 8.00 cm outside the paint layer: -7.27\cdot 10^7 N/C

Explanation:

a)

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

\int EdS = \frac{q}{\epsilon_0}

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

\epsilon_0 = 8.85\cdot 10^{-12}F/m is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius r as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

q=0

So we have

\int E dS=0

which means that the electric field inside the paint layer is zero.

b)

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

r=R=0.065 m

The area of the surface is

A=4\pi R^2

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}

The charge included within the sphere in this case is the charge on the paint layer, therefore

q=-17.0\mu C=-17.0\cdot 10^{-6}C

So, the electric field is:

E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C

where the negative sign means the direction of the field is inward, since the charge is negative.

c)

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m

Gauss theorem this time becomes

E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}

And the charge included within the sphere is again the charge on the paint layer,

q=-17.0\mu C=-17.0\cdot 10^{-6}C

Therefore, the electric field is

E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0
3 years ago
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