Which of the following are functions of both cells and living organisms?

What happens to the electronegativity of elements as you move from bottom left to upper right across the periodic table?

Paraphin [41]

It increases. As it moves it <span>increases while the movement is in process.

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3 0

3 years ago

A mole of a monatomic ideal gas at point 1 (101 kPa, 5 L) is expanded adiabatically until the volume is doubled at point 2. Then

Paha777 [63]

Answer:

(a). Check attachment.

(b). 280.305 J.

(c). 31.81 kpa; 38.26K.

(d). 24.05K.

(e). 24.05k; 40kpa.

(f). -138.6J.

Explanation:

(a). Kindly check the attached picture for the diagram showing the four process.

1 - 2 = adiabatic expansion process.

2 - 3 = Isochoric process.

3 - 4 = isothermal process.

4 - 1 = isochoric process.

(b). Recall that the process from 1 to is an adiabatic expansion process.

NB: b = 5/3 for a monoatomic gas.

Then, the workdone = (1/ 1 - 1.66) [ (p1 × v1^b)/ v2^b × v2 - (p1 × v1)].

= ( 1/ 1 - 5/3) [ (101 × 5^5/3) × 10^1 -5/3] - 101 × 5.

Thus, the workdone = 280.305 J.

(c). P2 = P1 × V1^b/ V2^b = 101 × 5^5/3/ 10^5/3 = 31.81 kpa.

T2 = P2 × V2/ R × 1 = 31.81 × 10/ 8.324 = 38.36k.

(d). The process 2 - 3 is an Isochoric process, then;

T3 = T2/P2 × P3 = 38.26/ 31.82 × 20 = 24.05K.

(e). The process 3 - 4 Is an isothermal process. Then, the temperature at 4 will be the same temperature at 3. Tus, we have the temperature; point 3 = point 4 = 24.05k.

The pressure can be determine as below;

P4 = P3 × V3/ V4 = 20 × 10/ 5 = 200/ 5 = 40 kpa.

(f) workdone = xRT ln( v4/v3) = 1 × 8.314 × 24.05 × ln (5/10) = - 138.6 J

6 0

3 years ago

What law describes light

Darina [25.2K]

The law of reflection

6 0

3 years ago

A piece of copper wire with thin insulation, 200 m long and 1.00 mm in diameter, is wound onto a plastic tube to form a long sol

bearhunter [10]

Answer:

N= 3

Explanation:

For this exercise we must use Faraday's law

E = - dФ / dt

Ф = B . A = B Acos θ

tje bold indicate vectors. As it indicates that the variation of the field is linear, we can approximate the derivatives

E = - A cos θ (B - B₀) / t

The angle enters the magnetic field and the normal to the area is zero

cos 0 = 1

A = π r²

In the length of the wire there are N turns each with a length L₀ = 2π r

L = N (2π r)

r = L / 2π N

we substitute

A = L² / (4π N²)

The magnetic field produced by a solenoid is

B = μ₀ N/L I

for which

B₀ = μ₀ N/L I

The final field is zero, because the current is zero

B = 0

We substitute

E = - (L² / 4π N²) (0 - μ₀ N/L I) / t

E = μ₀ L I / (4π N t)

N = μ₀ L I / (4π t E)

The electromotive force is E = 0.80 mV = 0.8 10⁻³ V

let's calculate

N = 4π 10⁻⁷ 200 1.60 / (4π 0.120 0.8 10⁻³)]

N = 320 10⁻⁷ / 9.6 10⁻⁶

N = 33.3 10⁻¹

N= 3

7 0

2 years ago

Exercises

Crank

$\\ \rm\Rrightarrow \dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$

$\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{1}{-10}+\dfrac{1}{38}$

$\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{-19+5}{190}$

$\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{-14}{190}$

$\\ \rm\Rrightarrow u=\dfrac{190}{-14}$

$\\ \rm\Rrightarrow u=13.6cm$

Real

5 0

2 years ago