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a weight is suspended from the ceiling by a spring (k = 20 ln/in) and is connected to the floor by a dashpot producing viscous damping. The damping force is 10 lb when the velocity of the dashpot plunger is 20 in/sec. The weight and plunger have W = 12 lb. What will be the frequency of the damped vibrations?

Answers

Answer 1

Answer:

Answer:

The frequency of the damped vibrations is 3.82 Hz.

Explanation:

Given that,

Spring constant = 20 lb/in

Damping force = 10 lb

Velocity = 20 in/sec

Weight = 12 lb

We need to calculate the damping constant

Using formula of damping force

$b* v=F_(d)$

$b=(F_(d))/(v)$

Put the value into the formula

$b =(10)/(20)$

$b=0.5\ lb-sec/in$

$b=0.5*12 =6\ lb-sec/ft$

We need to calculate the frequency

Using formula of angular frequency

$\omga=\sqrt{\omega_(0)^2-((b)/(2m))^2}$

$\omega=\sqrt{(k)/(m)-((b)/(2m))^2}$

Put the value into the formula

$\omega=\sqrt{(20*12*32)/(12)-((6*32)/(2*12))^2}$

$\omega=24\ rad/s$

We need to calculate the frequency of the damped vibrations

Using formula of frequency

$f=(\omega)/(2\pi)$

Put the value into the formula

$f=(24)/(2\pi)$

$f=3.82\ Hz$

Hence, The frequency of the damped vibrations is 3.82 Hz.

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Two charged particles attract each other with a force of magnitude F acting on each. If the charge of one is doubled and the distance separating the particles is also doubled, the force acting on each of the two particles has magnitude (a) F/2,
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Answers

Answer:

F/2

Explanation:

In the first case, the two charges are Q1 and Q2 and the distance between them is r. K is the Coulomb's constant

Hence;

F= KQ1Q2/r^2 ------(1)

Where the charge on Q1 is doubled and the distance separating the charges is also doubled;

F= K2Q1 Q2/(2r)^2

F2= 2KQ1Q2/4r^2 ----(2)

F2= F/2

Comparing (1) and (2)

The magnitude of force acting on each of the two particles is;

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What If? Fluoride ions (which have the same charge as an electron) are initially moving with the same speed as the electrons from part (a) through a different uniform electric field. The ions come to a stop in the same distance d. Let the mass of an ion be M and the mass of an electron be m. Find the ratio of the magnitude of electric field the ions travel through to the magnitude of the electric field found in part (a). (Use the following as necessary: d, K, m, M, and e for the charge of the electron.)

Answers

Answer:

E₁ / E₂ = M / m

Explanation:

Let the electric field be E₁ and E₂ for ions and electrons respectively .

Force on ions = E₁ e where e is charge on ions .

Acceleration on ions a = E₁ e / M . Let initial velocity of both be u . Final velocity v = 0

v² = u² - 2as

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u² = 2 x E₁ e d / M

Similarly for electrons

u² = 2 x E₂ e d / m

Hence

2 x E₁ e d / M = 2 x E₂ e d / m

E₁ / E₂ = M / m

Final answer:

The ratio of the magnitude of the electric field the ions travel through to the magnitude of the electric field found in part (a) is M/m.

Explanation:

The ratio of the magnitude of the electric field the ions travel through to the magnitude of the electric field found in part (a) can be determined using the concept of mechanical energy conservation. Since the ions come to a stop, their initial kinetic energy must be equal to the work done by the electric field on them. The work done is given by the equation:

Work = Change in kinetic energy

The change in kinetic energy can be calculated using the formula:

Change in kinetic energy = (1/2)Mv2 - (1/2)mv2

where M and m are the masses of the ions and electrons respectively, and v is their initial speed. Solving for the ratio, we get:

Ratio = (1/2)M/(1/2)m = M/m

So, the ratio of the magnitude of electric field the ions travel through to the magnitude of the electric field found in part (a) is M/m.

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Tonya picks up a leaf from the ground and holds it at arm’s length. She lets go, and the leaf falls to the ground. What force pulled the leaf to the ground?

Answers

Answer:

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Explanation:

Gravity is constantly pulling objects downward. Without it, everything would float out into space.

I hope this answer helps :)

Answer:

The answer for the given question above would be option C. GRAVITATIONAL FORCE. Based on the given scenario above of a leaf that falls to the ground when Tonya let it go, the force that pulled the leaf to the ground is the gravitational force. This kind of force is a force that attracts any object with mass.

Hope this helps!!!

Dawn is trying to find out how much weight she can push across the room. She is really trying to find her

Answers

Dawn is trying to figure out how much weight she can push with her strength, or what her maximum pushing force is, across the room. She could do an experiment to find out.

She must first prepare a testing space with a flat, smooth surface to reduce friction. She can then progressively add weights to a cart or other object and use all of her strength to try to push it across the room. She can determine her maximum pushing force by noting the heaviest weight she can move. For a variety of jobs, including moving furniture or participating in physical sports that call for pushing heavy things, this knowledge can be essential.

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Answer:

Muscular strength

Explanation:

She is testing her strength while pushing the weights

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Answers

The land of airplane gear of an airplane can be idealized as the spring-mass-damper system shown in fig. 3.52. if the runway surface is described

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