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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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An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr

Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

$F= mg$

Where, f = restoring force

$kx=mg$

$k=\dfrac{mg}{x}$

Put the value into the formula

$k=\dfrac{2.15\times9.8}{0.0895}$

$k=235.41\ N/m$

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Here, $v = A\omega$

$K.E=\dfrac{1}{2}m\times(A\omega)^2$

Here, $\omega=\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}kA^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2$

$K.E=0.06500\ J$

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0

3 years ago

A ball is thrown horizontally from the top of a building 0.10 km high. The ball strikes the ground at a point 65 m horizontally

aivan3 [116]

Answer:

option B

Explanation:

given,

height of building = 0.1 km

ball strikes horizontally to ground at = 65 m

speed at which the ball strike = ?

vertical velocity = 0 m/s

time at which the ball strike

$s = \dfrac{1}{2}gt^2$

$t = \sqrt{\dfrac{2s}{g}}$

$t = \sqrt{\dfrac{2\times 100}{9.8}}$

t = 4.53 s

vertical velocity at the time 4.53 s = g × t = 9.8 × 4.53 = 44.39 m/s

horizontal velocity = $\dfrac{65}{4.53}$ =14.35 m/s

speed of the ball = $\sqrt{44.39^2+14.35^2}$

= 46.65 m/s

hence, the speed of the ball just before it strike the ground = 47 m/s

The correct answer is option B

5 0

3 years ago

List the 3 types of collisions that occur when your vehicle hits an object.

koban [17]

These collisions are: "a Vehicle Collision, a Human Collision, Internal Collision." A vehicle collision is a collision that involves two or more vehicles and is when the vehicles collide against each other creating a unbalanced force since how the force comes from opposite directions. A human collision would involve a vehicle and a human which would also be a unbalanced force but the human wouldn't have much affect of it's speed. A internal collision is when something happens inside the vehicle which decreases, or increases the vehicles speed.

Hope this helps!

7 0

3 years ago

Why is Carbon Tetrahydride a covalent bond? plzzz helpppp!!!!!!

arlik [135]

Answer:

carbon has four unpaired electrons in its valence shell . hydrogen having one unpaired electron in its valence shell comes to bond with carbon by sharing a pair of electrons .since carbon needs 4 electrons to be stable, 4 hydrogen atoms take part in the bond . It's a covalent bond because the difference between the electronegativity of carbon and hydrogen is quite small .

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3 years ago

A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind

Marat540 [252]

Answer:

Part a)

$F_v = 4.28 N$