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Anastasy [175]
2 years ago
13

Hi, Please help me with the 2 questions, will mark and 5 stars and THANKS

Physics
2 answers:
Anit [1.1K]2 years ago
7 0

Both answers are going to be C

BlackZzzverrR [31]2 years ago
4 0

Answer: C and C

Explanation: C  since both parents contribute 50% of genetic information

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Earth's subsystems are the geosphere, atmosphere, hydrosphere, and biosphere. Energy and matter are transferred when Earth's sub
Leona [35]

Answer:

ok

Explanation:

The Earth is made of several subsystems or "spheres" that interact to form a complex and continuously changing whole called the Earth system. Scale

Processes operating in the Earth system take place on spatial scales varying from fractions of millimeters to thousands of kilometers, and on time scales that range from milliseconds to billions of years.

Examples of instantaneous - breathing; rotation of the Earth; earthquake

Examples of long term - making coal; plate tectonics

Cycles

The Earth system is characterized by numerous overlapping cycles in which matter is recycled over and over again. Cycles involve multiple spheres and systems interactions.

Examples of cycles: day and night; rock cycle; seasons

Energy

The Earth system is powered by energy from two major sources: the Sun and the planet's internal heat.

Humans and the Earth System

People are part of the Earth system and they impact and are impacted by its materials and processes.

6 0
2 years ago
A sample of monatomic ideal gas occupies 5.00 L at atmospheric pressure and 300 K (point A). It is warmed at constant volume to
leonid [27]

Answer:

(a) 0.203 moles

(b) 900 K

(c) 900 K

(d) 15 L

(e) A → B, W = 0, Q = Eint = 1,518.91596 J

B → C, W = Q ≈ 1668.69974 J Eint = 0 J

C → A, Q = -2,531.5266 J, W = -1,013.25 J, Eint = -1,518.91596 J

(g) ∑Q = 656.089 J, ∑W =  655.449 J, ∑Eint = 0 J

Explanation:

At point A

The volume of the gas, V₁ = 5.00 L

The pressure of the gas, P₁ = 1 atm

The temperature of the gas, T₁ = 300 K

At point B

The volume of the gas, V₂ = V₁ = 5.00 L

The pressure of the gas, P₂ = 3.00 atm

The temperature of the gas, T₂ = Not given

At point C

The volume of the gas, V₃ = Not given

The pressure of the gas, P₃ = 1 atm

The temperature of the gas, T₂ = T₃ = 300 K

(a) The ideal gas equation is given as follows;

P·V = n·R·T

Where;

P = The pressure of the gas

V = The volume of the gas

n = The number of moles present

R = The universal gas constant = 0.08205 L·atm·mol⁻¹·K⁻¹

n = PV/(R·T)

∴ The number of moles, n = 1 × 5/(0.08205 × 300) ≈ 0.203 moles

The number of moles in the sample, n ≈ 0.203 moles

(b) The process from points A to B is a constant volume process, therefore, we have, by Gay-Lussac's law;

P₁/T₁ = P₂/T₂

∴ T₂ = P₂·T₁/P₁

From which we get;

T₂ = 3.0 atm. × 300 K/(1.00 atm.) = 900 K

The temperature at point B, T₂ = 900 K

(c) The process from points B to C is a constant temperature process, therefore, T₃ = T₂ = 900 K

(d) For a constant temperature process, according to Boyle's law, we have;

P₂·V₂ = P₃·V₃

V₃ = P₂·V₂/P₃

∴ V₃ = 3.00 atm. × 5.00 L/(1.00 atm.) = 15 L

The volume at point C, V₃ = 15 L

(e) The process A → B, which is a constant volume process, can be carried out in a vessel with a fixed volume

The process B → C, which is a constant temperature process, can be carried out in an insulated adjustable vessel

The process C → A, which is a constant pressure process, can be carried out in an adjustable vessel with a fixed amount of force applied to the piston

(f) For A → B, W = 0,

Q = Eint = n·cv·(T₂ - T₁)

Cv for monoatomic gas = 3/2·R

∴ Q = 0.203 moles × 3/2×0.08205 L·atm·mol⁻¹·K⁻¹×(900 K - 300 K) = 1,518.91596 J

Q = Eint = 1,518.91596 J

For B → C, we have a constant temperature process

Q = n·R·T₂·㏑(V₃/V₂)

∴ Q = 0.203 moles × 0.08205 L·atm/(mol·K) × 900 K × ln(15 L/5.00 L) ≈ 1668.69974 J

Eint = 0

Q = W ≈ 1668.69974 J

For C → A, we have a constant pressure process

Q = n·Cp·(T₁ - T₃)

∴ Q = 0.203 moles × (5/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -2,531.5266 J

Q = -2,531.5266 J

W = P·(V₂ - V₁)

∴ W = 1.00 atm × (5.00 L - 15.00 L) = -1,013.25 J

W = -1,013.25 J

Eint = n·Cv·(T₁ - T₃)

Eint = 0.203 moles × (3/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -1,518.91596 J

Eint = -1,518.91596 J

(g) ∑Q = 1,518.91596 J + 1668.69974 J - 2,531.5266 J = 656.089 J

∑W = 0 + 1668.69974 J -1,013.25 J = 655.449 J

∑Eint = 1,518.91596 J + 0 -1,518.91596 J = 0 J

5 0
3 years ago
Help please !!!!<br> LIKE ASAP!
serg [7]
The first diagram is Nuclear Fission which is the splitting of a heavy, unstable nucleus into 2 lighter nuclei.

The second diagram is Nuclear Fusion which is the process where 2 light nuclei combine together releasing a lot of energy.
3 0
3 years ago
What are the output waveforms of the following waves, after passing through a transformer?
Ber [7]
The output waveforms after passing through the transformer actually depend on the type of transformer used. It could either be a step-up transformer (steps voltage up), or a step-down transformer (steps voltage down). Both transformers have an output voltage in a form of a sine wave.
8 0
3 years ago
Cables and conductors installed on the side of a pole or building shall be protected up to ___feet above finished grade.
Allisa [31]

Answer:

10 feet.

Explanation:

Conductors are considered outside the building when they installed. The requirement of NEC is above 10 feet for the clearance after the final grade for service conductors and cables where they are connected with the building, above the sidewalks and the to other areas which are accessible to the pedestrians.

8 0
3 years ago
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