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Advocard [28]
3 years ago
6

A student wants to demonstrate entropy using the songs on her portable music player. What should she do to demonstrate the lowes

t entropy? A. play a single song by a single artist B. play ONLY the classical songs on shuffle C. play all the songs in alphabetical order D. play all the songs on shuffle
Physics
1 answer:
antoniya [11.8K]3 years ago
4 0

Answer:  A. play a single song by a single artist

Explanation: Entropy is the measure of randomness or disorder of a system.

A system that has more random distribution is considered to have more entropy and a system with ordered arrangement is considered to have less entropy.

Playing a single song by a single artist has least randomness , and lowest disorder and thus has lowest entropy.

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Phase change occurs in which of the following A. A suspension. B boiling point. C An explosion D Mixing Point
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C I think not totally sure
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A jet liner must reach a speed of 82 m/s for takeoff. If the
SIZIF [17.4K]

Answer:

The acceleration that the jet liner that must have is 2.241 meters per square second.

Explanation:

Let suppose that the jet liner accelerates uniformly. From statement we know the initial (v_{o}) and final speeds (v_{f}), measured in meters per second, of the aircraft and likewise the runway length (d), measured in meters. The following kinematic equation is used to calculate the minimum acceleration needed (a), measured in meters per square second:

a = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot d}

If we know that v_{o} = 0\,\frac{m}{s}, v_{f} = 82\,\frac{m}{s} and d = 1500\,m, then the acceleration that the jet must have is:

a = \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (1500\,m)}

a = 2.241\,\frac{m}{s^{2}}

The acceleration that the jet liner that must have is 2.241 meters per square second.

3 0
2 years ago
how much force would be required to produce 88 j of work when pushing a box 1.1meters at an angle of 10 degrees?
ycow [4]

Answer:81.235N

Explanation:

Work=88J

theta=10°

distance=1.1 meters

work=force x cos(theta) x distance

88=force x cos10 x 1.1 cos10=0.9848

88=force x 0.9848 x 1.1

88=force x 1.08328

Divide both sides by 1.08328

88/1.08328=(force x 1.08328)/1.08328

81.235=force

Force=81.235

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3 years ago
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If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed
beks73 [17]

Answer:

v = 5.9 x 10⁷ m/s

Explanation:

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K.E = eV--------------- equation (1)

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V = Potential Difference = 9.9 KV = 9900 Volts

The kinetic energy in general is given as:

K.E = \frac{1}{2}mv^{2}\\--------- equation (2)

where,

m = mass of electron = 9.1 x 10⁻³¹ kg

v = speed of electron = ?

Therefore, comparing equation (1) and equation (2), we get:

\\\frac{1}{2}mv^{2} = eV\\\\\frac{1}{2}(9.1\ x\ 10^{-31}\ kg)v^{2} = (1.6\ x\ 10^{-19}\ C)(9900\ volts)\\\\v = \sqrt{34.81\ x\ 10^{14}} \\

<u>v = 5.9 x 10⁷ m/s</u>

8 0
3 years ago
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