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mariarad [96]
2 years ago
7

Which best describes the relationship between energy and entropy in the universe? for entropy to increase, energy must be added.

for entropy to decrease, energy must be added. for entropy to remain constant, energy must be added.
Physics
1 answer:
Gemiola [76]2 years ago
6 0

Answer:

Entropy is a measure of the order/disorder during the transformation of the state of a system and is defined as the total variation of energy at a defined temperature. From point of view of statistical mechanics, this variation of energy is generated from statistical transitions of the internal states of the system. In this sense, entropy can measure how easy it is to reach a defined state of the system. Now, imagine a text stream that arrives to you character by character in a screen. If the text is meaningless, then every character has the same probability of appearing to you and therefore the entropy is maximal because this disorder is maximal. If you want to transfer information, then you have to spend a little bit of energy in ordering the characters because this does not happen spontaneously. The final state of the system is more ordered in respect to the earlier one, so the entropy is less than the entropy of random text. This means that, if you want to reduce entropy in order to transfer information, then you must spend energy.

Explanation:

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A small compass is placed near a very large piece of ferromagnetic material. Before the compass is placed near the material, it
NeTakaya

Answer:

it will move towards the object's magnetic south

Explanation:

The compass pints towards the earth geographic north because the magnetic south of the earth's magnetic field is located in there, if you placed such compass neaar the piece of ferromagnetic material, the magnetic field produced by the magnet will make the compass needle point towards its south magnetic pole, in the same fashion as it points to the earth's magnetic south. It will point to the object's south pole because the magnetic field will be stronger than the earth's (which is weak) that is because of the way magnetism works, opposite poles are attracted and similar poles will tend to separate from each other

7 0
3 years ago
A 26.2-kg dog is running northward at 3.02 m/s, while a 5.30-kg cat is running eastward at 2.74 m/s. Their 65.1-kg owner has the
REY [17]

Answer:

Angle with the +x axis is θ = 79.599degree

Then the velocity of owner = 1.235m/s

Explanation:

Given that the mass of dog is m1 =26.2 kg

velocity of dog is u1 = 3.02 m/s (north)

mass of cat is m2 = 5.3 kg

velocity is u2 = 2.74 m/s (east )

Mass of owner is M = 65.1 kg

Consider the east direction along +x axis andnorth along +y

momentum of dog is Py = m1 x u1

= 79.124 kg.m/s (j)

momentum of cat is Px = m2 x u2

= 14.522 kg.m/s (i)

Then the net magnitude of momentum is P = (Px2 + Py2)1/2

= 80.445

Angle with the +x axis is θ =tan-1(Py / Px ) = 79.599 degree

Then the velocity of owner is v = P / M = 1.235 m/s

3 0
3 years ago
A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A
Y_Kistochka [10]

Answer:

The tensions in T_{BC} is approximately 4,934.2 lb and the tension in T_{BD} is approximately  6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = T_{BC}

The tension in rope segment BD = T_{BD}

The tension in rope segment AB = T_{AB} = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

T_{BC} × cos(26.0°) + T_{BD} × cos(21.0°) = T_{AB} = 1200 lb

T_{BC} × cos(26.0°) + T_{BD} × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

T_{BC} × sin(26.0°) + T_{BD} × sin(21.0°) = 0...........................(2)

Which gives;

T_{BC} × sin(26.0°) = - T_{BD} × sin(21.0°)

T_{BC} = - T_{BD} × sin(21.0°)/(sin(26.0°))  ≈ - T_{BD} × 0.8175

Substituting the value of, T_{BC}, in equation (1), gives;

- T_{BD} × 0.8175 × cos(26.0°) + T_{BD} × cos(21.0°) = 1200 lb

- T_{BD} × 0.7348  + T_{BD} ×0.9336 = 1200 lb

T_{BD} ×0.1988 = 1200 lb

T_{BD} ≈ 1200 lb/0.1988 = 6,035.6938 lb

T_{BD} ≈ 6,035.6938 lb

T_{BC} ≈ - T_{BD} × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

T_{BC} ≈ -4934.1733 lb

From which we have;

The tensions in T_{BC} ≈ -4934.2 lb and  T_{BD} ≈ 6,035.7 lb.

8 0
3 years ago
Write down the boiling point and freezing point of mercury an alcohol​
nekit [7.7K]
Mercury has a high boiling point of 357 degrees C.
Mercury has a freezing point of −39 degrees C.
8 0
3 years ago
What is the magnitude of the gravitational force acting on the earth due to the sun?
expeople1 [14]

Answer: 3.524(10)^{22}N

Explanation:

According to Newton's law of Gravitation, the force F exerted between two bodies of masses m1 and m2  and separated by a distance r  is equal to the product of their masses and inversely proportional to the square of the distance:

F=G\frac{(m1)(m2)}{r^2}   (1)

Where:

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}  

m1=1.99(10)^{30}kg is the mass of the Sun

m2=5.972(10)^{24}kg is the mass of the Earth

r=1.50(10)^{11}m  is the distance between the Sun and the Earth

Substituting the values in (1):

F=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(1.99(10)^{30}kg)(5.972(10)^{24}kg)}{(1.50(10)^{11}m)^2}   (2)

Finally:

F=3.524(10)^{22}N   This is the gravitational force acting on the earth due to the sun

3 0
3 years ago
Read 2 more answers
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