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alexandr402 [8]
2 years ago
8

The motion of a particle connected to a spring is described by x = 10 sin (pi*t). At

Physics
1 answer:
Gekata [30.6K]2 years ago
6 0

To solve this problem we will first apply the principle of energy conservation, for which the elastic potential energy must be the same as the elastic kinetic energy of the simple harmonic movement of the system.

From this conservation it will be possible to find in total terms the total displacement of the system and thus replace it in the given function to find the time.

The potential energy stored would be

PE= \frac{1}{2} kx^2

The kinetic energy would be

KE = \frac{1}{2}m\omega (A^2-x^2)

Here,

A = 10m

\omega = \pi rad/s

Now assuming the planned conservation of energy

PE = KE

\frac{1}{2}kx^2 =\frac{1}{2} m\omega^2 (A^2-x^2)

\frac{1}{2}kx^2 = \frac{1}{2} m(x\frac{k}{m})(A^2-x^2)

x^2 = A^2 -x^2

2x^2 = A^2

x = \frac{A}{\sqrt{2}}

x = \frac{10}{\sqrt{2}}

So now using the equation previously given to describe the motion of the particle we have to

\frac{10}{\sqrt{2}} = 10 sin(xt)

\frac{1}{\sqrt{2}} = sin (xt)

sin^{-1} (\frac{1}{\sqrt{2}} ) = \pi t

\frac{\pi}{4} = \pi t

t = \frac{1}{4}s= 0.25s

Therefore the correct option is B.

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A rocket moves upward, starting from rest with an acceleration of +29.4 for 3.98 s. it runs out of fuel at the end of the 3.98 s
topjm [15]
U = 0, initial upward speed
a = 29.4 m/s², acceleration up to 3.98 s
a = -9.8 m/s², acceleration after 3.98s

Let h₁ =  the height at time t, for t ≤ 3.98 s
Let h₂ =  the height at time t > 3.98 s

Motion for  t ≤ 3.98 s:
h₁ = (1/2)*(29.4 m/s²)*(3.98 s)² = 232.854 m
Calculate the upward velocity at t = 3.98 s
v₁ = (29.4 m/s²)*(3.98 s) = 117.012 m/s

Motion for t  > 3.98 s
At maximum height, the upward velocity is zero.
Calculate the extra distance traveled before the velocity is zero.
(117.012 m/s)² + 2*(-9.8 m/s²)*(h₂ m) = 0
h₂ = 698.562 m

The total height is
h₁ + h₂ = 232.854 + 698.562 = 931.416 m

Answer: 931.4 m (nearest tenth)

6 0
2 years ago
Read 2 more answers
true or false:acceleration toward the center of a curved or circular path is called gravitational acceleration.
nalin [4]
Nope. It's called 'centripetal' acceleration. The force that created it MAY be gravitational, but it doesn't have to be. For things on the surface of the Earth moving in circles, it's never gravity.
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A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate
diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

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Where,

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T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

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According to the data given we have to,

Q_{source} = 200000Btu

T_{source} = 1500R

Q_{sink} = 100000Btu

T_{sink} = 600R

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\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}

\Delta S_{sink} = \frac{100000}{600}

\Delta S_{sink} = 166.67Btu/R

On the other hand,

\Delta S_{source} = \frac{Q_{source}}{T_{source}}

\Delta S_{source} = \frac{-200000}{1500}

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S = \Delta S_{source}+\Delta S_{sink}

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PART B)

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4 0
2 years ago
Liang is working with an electrical circuit. She replaces a straight electrical wire with a coiled wire. What is Liang most like
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3 0
2 years ago
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erik [133]

Answer:

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6 0
2 years ago
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