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2 years ago

the amplitude of an oscillator decreases to 36.8% of its initial value in 10.0 s. what is the value of the time constant

Physics

1 answer:

iragen [17]2 years ago

4 0

Answer:

τ = 5 s

Explanation:

When a vibrating body is damped. Its amplitude starts to decrease. This decrement is exponential. And it is given as follows:

$X = X_{0}e^{-\frac{t}{2\tau}$

where,

τ = Time Constant = ?

X = Instantaneous value of amplitude

X₀ = Initial Value of amplitude

t = time interval = 10 s

The ratio of decrement is given as:

$\frac{X}{X_0} = 36.8\% = 0.368$

therefore, using these values, we get:

$\frac{X}{X_{0}} = 0.368 = e^{\frac{10\ s}{2\tau}}$

Taking natural log (ln) on both sides, we get:

$ln(0.368) = \frac{10\ s}{2\tau}\\\\\tau = \frac{10\ s}{2ln(0.368)}$

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I believe the answer is D. phase changes. The two level portions represents change of state that does not involve change in temperature (at a constant temperature). The first level represents a change of solid to liquid;p where the ice melts and becomes water by gaining the latent heat of fusion, while the other level represents a change of state from liquid to gas; the water changes to steam (water vapor) by gaining the latent heat of vaporization.

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2 years ago

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un tanque de gasolina de 40 litros fue llenado por la noche, cuando la temperatura era de 68 grados farenheit. Al dia siguiente

Sedaia [141]

Answer:

Volume of gasoline that expands and spills out is 1.33 ltr

Explanation:

As we know that when temperature of the liquid is increased then its volume will expand and it is given as

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here we know that

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volume expansion coefficient of the gasoline is given as

$\gamma = 950 × 10^{–6}$

change in temperature is given as

$\Delta T = (131 - 68) \times \frac{5}{9}$

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$\Delta V = 40(950 \times 10^{-6})(35)$

$\Delta V = 1.33 Ltr$

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2 years ago

In the equation vx^2=v0x^2+2ax(x-x0) what does the terms vx, v0x, x, and x0 stand for respectively?

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B. velocity at position x, velocity at position x=0, position x, and the original position

In the equation

$v_{x}^{2}$ = $v_{ox}^{2}$ +2 a x (x - x₀)

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$v_{ox}$ = velocity at position "x = 0 "

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John is carrying a shovelful of snow. The center of mass of the 3.00 kg of snow he is holding is 15.0 cm from the end of the sho

Andru [333]

Answer:

James is correct here as the force of hand pushing upwards is always more than the force of hand pushing down

Explanation:

Here we know that one hand is pushing up at some distance midway while other hand is balancing the weight by applying a force downwards

so here we can say

Upwards force = downwards Force + weight of snow

while if we find the other force which is acting downwards

then for that force we can say that net torque must be balanced

so here we have

$F_{down} L_1 = W_{snow} L_2$

so here we have

$F_{down} = \frac{L_2}{L_1} (W_{snow})$

so here we can say that upward force by which we push up is always more than the downwards force

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2 years ago

A 60.0-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.760 and 0.410,

Alik [6]

The horizontal pushing force required to just start the crate moving is 447 N.

The horizontal pushing force required to slide the crate across the dock at a constant speed is 241 N.

<u>Explanation</u>:

By the definition of the coefficient of static friction we have:

μ $_{s}$ = $\frac{F_{appl} }{W}= \frac{F_{s} }{N}$ ,

where, $F_{appl}$ is the horizontal pushing force,

W = mg is the weight of the crate directed downward,

$F_{s}$ is the static friction force-directed opposite to the horizontal pushing force and equal to it,

N is the force of reaction directed upward and equal to the weight of the crate.

From this formula we can find the horizontal pushing force required to just start the crate moving:

$F_{appl} = F_{s} = u_{s}N = u_{s}mg$

= 0.760 $\times$ 60 kg $\times$ 9.8 m / s^2

= 447 N.

By the definition of the coefficient of kinetic friction we have:

u $_{k} = \frac{F_{appl} }{W} = \frac{F_{k} }{N}$ ,

where, $F_{appl}$ is the horizontal pushing force,

W = mg is the weight of the crate directed downward,

$F_{k}$ is the kinetic friction force-directed opposite to the horizontal pushing force and equal to it,

N is the force of reaction directed upward and equal to the weight of the crate.

From this formula we can find the horizontal pushing force required to slide the crate across the dock at a constant speed:

$F_{appl} = F_{k} = u_{k}N = u_{k}mg$

= 0.410 $\times$ 60 $\times$ 9.8

= 241 N.

The horizontal pushing force required to just start the crate moving is 447 N.

The horizontal pushing force required to slide the crate across the dock at a constant speed is 241 N.

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2 years ago