A mass weighing pounds is attached to a spring whose constant is lb/ft. The medium offers a damping force that is numerically equal to the instantaneous velocity. The mass is initially released from a point foot above the equilibrium position with a downward velocity of ft/s. Determine the time at which the mass passes through the equilibrium position. Find the time at which the mass attains its extreme displacement from the equilibrium position. What is the position of the mass at this instant

Answers

Answer 1

Answer:

hello your question has some missing values attached below is the complete question with the missing values

answer :

a) 0.083 secs

b) 0.33 secs

c)  3e^-4/3

Explanation:

Given that

g = 32 ft/s^2 ,  spring constant ( k ) = 2 Ib/ft

initial displacement = 1 ft above equilibrium

mass = weight / g = 4/32 = 1/8

damping force = instanteous velocity  hence  β = 1

a)Calculate the time at which the mass passes through the equilibrium position.

time mass passes through equilibrium = 1/12 seconds = 0.083

b) Calculate the time at which the mass attains its extreme displacement

time when mass attains extreme displacement = 1/3 seconds = 0.33 secs

c) What is the position of the mass at this instant

position = 3e^-4/3

attached below is the detailed solution to the given problem

A Mass Weighing Pounds Is Attached To A Spring Whose Constant Is Lb/ft. The Medium Offers A Damping Force
A Mass Weighing Pounds Is Attached To A Spring Whose Constant Is Lb/ft. The Medium Offers A Damping Force
A Mass Weighing Pounds Is Attached To A Spring Whose Constant Is Lb/ft. The Medium Offers A Damping Force

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Answer:

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Answer:

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Explanation:

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Answers

Answer:

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Explanation:

From the question,

Using

a = (v-u)/t.................... Equation 1

Where a = accelartion of the bicycle, v = Final velocity, u = initial velocity, t = time.

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Substitute these values into equation 1

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Hence the acceleration of the bicycle is 1 m/s²

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Answers

Answer:

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Explanation:

The resistance of a conductor in terms of its physical dimensions can be given by the following formula:

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using equation (1):

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In the 1500's Galileo experimented and discovered many things. One of his famous experiments allowed him to discover that blank when you drop them.​

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Answer:

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Explanation:

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Answer:

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Explanation:

Answer:

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Answer:

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Explanation:

From the question,

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Answer:

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mark me as brainlist ez

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Answer:

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Answer:

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Answer:

Explanation:

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A boy who is riding his bicycle, moves with an initial velocity of 5 m/s. ten second later, he is moving at 15 m/s. what is his acceleration

Answers

[tex]\Large {{ \sf {Question :}}} [/tex]

A boy who is riding his bicycle, moves with an initial velocity of 5 m/s. Ten second later, he is moving at 15 m/s. What is his acceleration?

[tex]\Large {{ \sf {Given :}}} [/tex]

Initial Velocity (u) - 5 m/sFinal Velocity (v) - 15 m/sTime (t) - 10 sec

[tex]\Large {{ \sf {Formulae :}}} [/tex]

If the velocity of an object changes from an initial value u to the final value v in time t, the acceleration a is, [tex]a \: = \frac{v - u}{t} [/tex][tex]\Large {{ \sf {Step-by-step explanation :}}} [/tex]

[tex]a \: = \frac{v - u}{t} \\ or \: \: a = \frac{(15 - 5)}{10} m \: s^{ - 2} \\ or \: \: a \: = \frac{10}{10}m \: s^{ - 2} \\ or \: \: a = 1m \: s^{ - 2} [/tex]

[tex]\Large {{ \sf {Answer :}}} [/tex]

His acceleration is [tex]1m \: s^{ - 2} [/tex]

can someone help me please? thank you

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....................

If spring has a spring constant of 500 N/m and is stretched .50 meters,how much energy is stored in the spring ((show work for full credit, show equation))

Answers

Answer:

The energy stored is: 62.5 Joules

Explanation:

Given

[tex]k = 500N/m[/tex] --- spring constant

[tex]x = 0.5m[/tex] --- stretch

Required

The amount of energy

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[tex]U = \frac{1}{2} kx^2[/tex]

[tex]U = \frac{1}{2} * 500N/m * (0.5m)^2[/tex]

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Please please please please please please please please please please please please please please please please please please please

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c i think & mark as brain list?

PLEASE HELP! A 0.423 kg object carries a +11.5 uc
charge. It is 0.925 m from a -7.55 C
charge. What is the magnitude of the
object's acceleration?
(u stands for micro. Remember, magnitudes
are positive.)
[?] m/s2

Answers

Answer:

2.15 m/s²

Explanation:

We'll begin by calculating the force of attraction between two charges. This can be obtained as follow:

Charge of 1st object (q₁) = +11.5 μC = +11.5×10¯⁶ C

Charge of 2nd object (q₂) = –7.55 μC = –7.55×10¯⁶ C

Electrical constant (K) = 9×10⁹ Nm²/C²

Distance apart (r) = 0.925 m

Force (F) =?

F = Kq₁q₂ / r²

F = 9×10⁹ × 11.5×10¯⁶ × 7.55×10¯⁶/ 0.925²

F = 0.781425 / 0.855625

F = 0.91 N

Finally, we shall determine the acceleration of the object. This can be obtained as follow:

Mass of object (m) = 0.423 Kg

Force (F) = 0.91 N

Acceleration (a) =?

F = ma

0.91 = 0.423 × a

Divide both side by 0.423

a = 0.91 / 0.423

a = 2.15 m/s²

Thus, the magnitude of the object's acceleration is 2.15 m/s²

Answer:

2.15

Explanation:

Acellus

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A 12 N of force pulling left while a 22 N of force pulls right.

A 6 N of force pulling left while a 10 N of force pulls right.

A 2 N force and a 10 N force pulling left while a 12 N force pulls right.

A 22 N force pulling left while a 6 N force and a 12 N force pulls right.

Answers

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A 2 N force pulling and a 10 N force pulling left while a 12N force pulls right.

A cog system on the beginning segment of a roller coaster needs to get 29 occupied cars up a 120-m vertical rise over a time interval of 60 s. Each car experiences a gravitational force of 5900 N. The cars start at rest and end up moving at 0.50 m/s.

Answers

Answer:

Hello your question has some missing part attached below is the missing part

How much work is done by the cog system on the cars?

answer : 2.053 * 10^7 J

Explanation:

Total weight of car ( wT )  = 5900 * 29 = 171100 N

mass of car ( mT ) = wT / g  = 171100 / 9.81 = 17459.18 kg

Hence work done by the cog system on the cars

W = ( KE[tex]_{2}[/tex] + PE[tex]_{2}[/tex] ) - ( KE[tex]_{1}[/tex] + PE[tex]_{1}[/tex] )

    = ( 1/2 mv^2 + mgh ) - ( 0 )

    = ( 1/2 * 17459.18 * 0.50^2) + ( 17459.18 * 9.81 * 120 )

    = 2.053 * 10^7 J

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