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natita [175]
3 years ago
5

According to the Second Law of Thermodynamics, a. any process during which the entropy of the universe increases will be product

-favored. b. heat will always be transferred from a cooler object to a warmer object. c. any process during which the entropy of the system increases will be product-favored. d. any process which is endothermic will be product-favored. e. the total amount of matter in the universe remains constant.
Physics
1 answer:
Iteru [2.4K]3 years ago
4 0

Answer:

The correct option is;

a. Any process in which the entropy of the universe increases will be product-favored

Explanation:

According to the second law of thermodynamics, the change in entropy of a closed system with time is always positive. That is the entropy of the entire universe, considered as an isolated system, always increases with time, hence the entropy change in the universe will always be positive.

\Delta S_{universe} = \Delta S_{system}  + \Delta S_{surroundings} >0

Therefore, any process in which the entropy of the universe increases will be product favored.

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The answer is C guide and inspire good conduct
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2 years ago
The great limestones caverns were formed by dripping water. If water droplets of 10 ml fall from a height of 5 m at a rate of 10
loris [4]

The average force of the water droplets is the force given by the impact

per second of the droplets on the limestone floor.

  • The average force exerted on the limestone floor is approximately <u>1.6013 × 10⁻² N</u>

Reasons:

The given parameters are;

Volume of a droplet = 10 ml = 1 × 10⁻⁵ m³

Height from which the water falls, <em>h </em>= 5 meters

Rate at which the water falls = 10 per minute

Required:

The average force exerted on the floor by the water droplets.

Solution:

According to Newton's Second Law of motion, we have;

Force = Rate of change of momentum

Momentum = Mass × Velocity

Mass of a droplet of water = Volume × Density

Density of water = 997 kg/m³

Mass of a droplet = 1 × 10⁻⁵ m³ × 997 kg/m³ = 0.00997 kg

The velocity just before the droplet reaches the ground, v = √(2·g·h)

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Which gives;

v = √(2 × 9.81 m/s² × 5 m) ≈ 9.905 m/s

The rate of change in momentum per minute = 1

Therefore;

\displaystyle The \ rate \ of \ change \ in \ momentum = Average \ force = \mathbf{\frac{\Delta Momentum }{\Delta Time}}

ΔMomentum = Mass × ΔVelocity

Considering the 10 drops per minute, we have;

ΔMomentum = 10 × 0.0097 kg × 9.905 m/s = 0.960785 kg·m/s

ΔTime = 1 minute = 60 seconds

Therefore;

\displaystyle Average \ force, \, F_{ave}  \frac{0.960785 \, kg\cdot m/s }{60 \, s} \approx =\mathbf{1.6013 \times 10^{-2} \, N}

  • The average force exerted on the limestone floor by the droplets of water is F_{ave} ≈ <u>1.6013 × 10⁻² N</u>

Learn more about Newton's Second Law of motion and force exerted water here:

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3 0
2 years ago
A 15.0 kg turntable with a radius of 25 cm is covered with a uniform layer of dry ice that has a mass of 9.0 kg. The angular spe
liubo4ka [24]

Answer:

 ω₂=1.20

Explanation:

Given that

mass of the turn table ,M= 15 kg

mass of the ice ,m= 9 kg

radius ,r= 25 cm

Initial angular speed ,ω₁ = 0.75 rad/s

Initial mass moment of inertia

I_1=\dfrac{M+m}{2}r^2

I_1=\dfrac{15+9}{2}\times 0.25^2\ kg.m^2

I_1=0.75\ kg.m^2

Final mass moment of inertia

I_2=\dfrac{M}{2}r^2

I_2=\dfrac{15}{2}\times 0.25^2\ kg.m^2

I_2=0.468\ kg.m^2

Lets take final speed of the turn table after ice evaporated =ω₂ rad/s

Now by conservation angular momentum

I₁ ω₁ =ω₂ I₂

\omega_2=\dfrac{0.75\times 0.75}{0.468}\ rad/s

ω₂=1.20

7 0
3 years ago
Your skeleton. It is there for support, protection, and movement. But you could not move your bones without the help of your ___
lana66690 [7]
I believe the answer is D.) the muscular system
7 0
3 years ago
Read 2 more answers
How much energy is required to raise the temperature of 50.0 grams of water 10.0 degree C? (Explain yourself answer in joules!)
maria [59]

The amount of energy needed is 2093 J

Explanation:

The amount of energy needed to increase the temperature of a substance by \Delta T is given by the equation

Q=mC\Delta T

where

m is the mass of the substance

C is its specific heat capacity

\Delta T is the increase in temperature

For the water in this problem, we have

m = 50.0 g = 0.050 kg

C=4186 J/g^{\circ}C (specific heat capacity of water)

\Delta T=10.0^{\circ}C

Therefore, the amount of energy needed is

Q=(0.050)(4186)(10)=2093 J

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