A 12-kg block on a horizontal frictionless surface is attached to a light
spring (force constant = 0.80 kN/m). The block is initially at rest at its
equilibrium position when a force (magnitude P = 80 N) acting parallel to
the surface is applied to the block, as shown. What is the speed of the
block when it is 13 cm from its equilibrium position?"


Answers

Answer 1

The speed of the block at the displacement from the equilibrium position is 1.062 m/s.

The given parameters:

Mass of the block, m = 12 kgSpring constant, k = 0.8 kN/mExtension of the spring, x = 13 cm = 0.13 mApplied parallel force, F = 80 N

The speed of the block is calculated by applying the principle of conservation of mechanical energy as shown below;

[tex]\frac{1}{2} mv^2 = \frac{1}{2}kx^2\\\\mv^2 = kx^2\\\\v^2 = \frac{kx^2}{m} \\\\v = \sqrt{\frac{kx^2}{m} } \\\\v = \sqrt{\frac{800 \times 0.13^2}{12} } \\\\v = 1.062 \ m/s[/tex]

Thus, the speed of the block at the displacement from the equilibrium position is 1.062 m/s.

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Answer 2

Answer:

The speed of the block at the displacement from the equilibrium position is 1.1266 m/s.

Step-by-step explanation:

Solution :

Using principle of conservation of mechanical energy formula to find the speed of the block :

[tex]\begin{gathered} \longrightarrow{\pmb{\sf{\frac{1}{2} mv^2 = \frac{1}{2}kx^2}}}\end{gathered}[/tex]

»» m = Mass of the block, »» k = Spring constant,»» x = Extension of the spring»» F = Applied parallel force

As per given data information in the question we have :

✧ Mass of the block = 12 kg✧ Spring constant = 0.8 kN/m✧ Extension of the spring = 0.13 m✧ Applied parallel force = 80 N

Substituting all the given values in the formula to find the speed of the block

[tex]\longrightarrow{\sf{ \: \:\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: \cancel{\dfrac{1}{2}}mv^2 = \cancel{\dfrac{1}{2}}kx^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: mv^2 = kx^2}}[/tex]

[tex]\longrightarrow{\sf{ \: \: v^2 = \dfrac{kx^2}{m}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: \sqrt{{v}^{2} } = \sqrt{ \dfrac{kx^2}{m}}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: v = \sqrt{ \dfrac{kx^2}{m}}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: v = \sqrt{ \dfrac{800 \times {0.13}^{2}}{12}}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: v = \sqrt{ \dfrac{800 \times {0.13} \times 0.13}{12}}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: v = \sqrt{ \dfrac{800 \times 0.0169}{12}}}}[/tex]

[tex]\longrightarrow{\sf{ \: \: v = \sqrt{ \dfrac{13.52}{12}}}}[/tex]

[tex]{\star{\underline{\boxed{\rm{\red{ v \approx 1.1266 \: m/s}}}}}}[/tex]

Hence, the speed of block is 1.1266 m/s.

[tex] \rule{300}{1.5}[/tex]


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How many protons, electrons and nurturing does krypton have NEED HELP ASAP THANK YOU

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Answer:

Krypton – Mass Number – Neutron Number – Kr 2020-11-21 by Nick Connor Krypton is a chemical element with atomic number 36 which means there are 36 protons and 36 electrons in the atomic structure. The chemical symbol for Krypton is Kr.

Atomic Number: 36

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imageshare.best/image.php?id=JS9NDO

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Which feature of a balanced chemical equation demonstrates the law of
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Answers

Answer:

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Explanation:

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Hope this helps you

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Answers

Answer:

The one falling from the greatest height will have the greatest speed.

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Check:

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A heat engine is a device that uses to produce useful work.

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A child on ice skates is given a small push from behind by a parent. There is an unbalanced force on the child, and the child’s motion will change direction or increase speed.

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A child on ice skates is given a small push from behind by a parent. There is an unbalanced force on the child, and the child’s motion will change direction or increase speed.

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Answer:  Your answer Is A)

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Answers

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Answers

The mass of  potassium nitrate (KNO₃) crystals that will be separated is calculated as 6.25 g.

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Answers

Remain the same.

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Answers

Answer:

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Cool. What’s the Question?

Explanation:

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Answers

Answer:

Explanation:

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The security alarm on a parked car goes off and produces a frequency of 769 Hz. The speed of sound is 343 m/s. As you drive toward this parked car, pass it, and drive away, you observe the frequency to change by 69.5 Hz. At what speed are you driving

Answers

Answer:

Explanation:

ASSUMING your speed is constant

f₀ = f(v + vo)/(v + vs)

   Δf = f approach - f depart

69.5 = (769(343 + vo)/(343 + 0)) - (769(343 - vo)/(343 + 0))

69.5 = 769(2vo/343)

  vo = 15.5 m/s

The speed of car driving is 15.5 m/s as the car is parked and drive away.

What is speed?

Speed is defined as a measurement of the length of time it takes for an object to travel a certain distance. You can determine an object's speed if you know how far it moves in a given amount of time. Time does not move, hence there is no concept of a speed of time. Time refers to how we move through the temporal realm. Speed is a unit of measurement for how quickly something is moving. A change in velocity results in a change in speed.

To calculate the speed we use the formula

f₀ = f (v + vo) / (v + vs)

Δf = f approach - f depart

69.5 = (769(343 + vo) / (343 + 0)) - (769(343 - vo)/(343 + 0))

69.5 = 769(2vo/343)

vo = 15.5 m/s

Thus, the speed of car driving is 15.5 m/s as the car is parked and drive away.

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For northern hemisphere observers, which celestial object would be above the horizon for the greatest
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Answers

Answer:

Explanation:

For a person at about 20° North latitude, an object 70° above the celestial equator would never set. It's arc path would touch the horizon be never sink below it. Observers north of 20° see it all night. Observers south of 20° an object 70° above the celestial equator would spend the greatest amount of time above the horizon.

For southern hemisphere observers, the object 40" below the celestial equator will spend the most time above the horizon. Nearly 12 hours per day. Did you mean 40°? 40 seconds is very close to the equator itself. However, the result is the same.

For northern hemisphere observers, the celestial object that is 70° above the celestial equator would be above the horizon for the greatest amount of time.

What is the equator?

The Equator is an imaginary line passing through the middle of a globe. It is equidistant from the North Pole and the South Pole, Its is a horizontal line residing at 0 degrees latitude.

For northern hemisphere observers, the celestial object that is 70° above the celestial equator would be above the horizon for the greatest amount of time.

One that is 40" below the celestial equator would be above the horizon for the greatest amount of time for southern hemisphere observers.

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PLEASE ANSWER THIS, I'LL MARK YOU AS BRAINLIEST

Carlito was observing an an that crawled along a table. With a piece of chalk, he followed his path. He determined the ant’s displacements by using a ruler and protractor. The displacements were as follows: 2 cm east, 3.5 cm, 32° north of east and 2.4 cm, 22° west of north. Find the resultant vector using graphical method.​

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Answer:

this is the answer

Explanation:

hope it's clear

pls Mark me as the brainliest pls

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