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uranmaximum [27]
2 years ago
9

A 16 lb weight stretches a spring 6 inches in equilibrium. It is attached to a damping mechanism with constant c. Find all value

s of c such that the free vibration of the weight has infinitely many oscillations.
Physics
1 answer:
arlik [135]2 years ago
6 0

Answer:

-8

Explanation:

Given data:

Weight W = 16 lb

length l = 6/12 = 0.5 ft

hence, spring constant  k = W/l = 16/0.5 = 32 lb/ft

The equation of motion of spring is

mx''+cx+kx=0\\0.5x''+cx'+32x=0

the auxiliary equation can be written as

0.5r^2+cr+32=0\\r^2+2cr+64=0

The discriminate of equation is

D =(2c)^2+-4(1)(64)\\=4(c^2-64)

To get the value of the damping constant,

D

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It is known that heat is added to a gas in sealed container. The container is fitted with a moveable piston.
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Answer:

d. Not enough information is given to answer this question.

Explanation:

From first law of thermodynamics

Q= W + ΔU

Q=Heat  ,W= Work , ΔU=Change in internal energy

If work done by the gas :

It means that W and Q both are positive

Q- W = ΔU

Ii Q > W ,then temperature of the gas will increase.

If  Q< W  ,Then temperature of the gas will decreases.

If work done on the gas:

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There are three cases because they did not give any information about the work.That is why option d is correct.

3 0
3 years ago
A substance has a half-life of 10,000 years and an initial mass of 1,000 grams. How many years will pass before only 250 grams o
solmaris [256]
<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ After 10,000 years the mass will be:

1,000 / 2 = 500

After 20,000 years the mass will be:

500/2 = 250

As you can see, the correct answer would be A. 20,000 years

<h3><u>✽</u></h3>

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6 0
3 years ago
Read 2 more answers
Electric field lines always begin at _______ charges (or at infinity) and end at _______ charges (or at infinity). One could als
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Electric field lines always begin at positive charges (or at infinity) and end at negative charges (or at infinity).

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6 0
3 years ago
A torsion-bar spring consists of a prismatic bar, usually of round cross section, that is twisted at one end and held fast at th
ra1l [238]

Answer:

d₁ = 0.29 in

d₂ = 0.505 in

Explanation:

Given:

T = 1500 lbf in

L = 10 in

x = 0.5 L = 5 in

T_{1} =\frac{T(L-x)}{L} =\frac{1500*(10-5)}{10} =750lbfin

First case: T = T₁ + T₂

T₂ = T - T₁ = 1500 - 750 = 750 lbf in

If the shafts are in series:

θ = θ₁ + θ₂

θ = ((T₁ * L₁)/GJ) + ((T₂ * L₂)/GJ)

Second case: If d₁ ≠ d₂

θ = ((T₁ * L₁)/GJ₁) + ((T₂ * L₂)/GJ₂) = 0 (eq. 1)

t₁ = t₂

\frac{16T_{1} }{\pi d_{1}^{3}  } =\frac{16T_{2} }{\pi d_{2}^{3}  } (eq. 2)

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θ₁ first case = θ₁ second case

Replacing:

\frac{750*5}{G(\frac{\pi }{32})*0.5^{4}  } =\frac{T_{1}*3.7 }{G(\frac{\pi }{32})*d_{1} ^{4}  }\\T_{1} =16216d_{1} ^{4}

The same way to θ₂:

\frac{750*5}{G(\frac{\pi }{32})*0.5^{4}  } =\frac{T_{2}*6.3 }{G(\frac{\pi }{32})*d_{2} ^{4}  } \\T_{2} =9523.8d_{2} ^{4}

From equation 2, we have:

d₁ = 0.587 * d₂

From equation 3, we have:

d₂ = 0.505 in

d₁ = 0.29 in

7 0
3 years ago
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