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Nitrogen (N2) undergoes an internally reversible process from 6 bar, 247°C during which pν1.2 = constant. The initial volume is 0.1 m3 and the work for the process is 50 kJ. Assuming ideal gas behavior, and neglecting kinetic and potential energy effects, determine heat transfer, in kJ, and the entropy change, in kJ/K.

Answers

Answer 1

Answer:

Answer:

$Q=25\ kJ$

$\Delta s= 0.2885 J.K^(-1)$

Explanation:

Given:

Initial pressure of nitrogen, $P_1=6* 10^5\ Pa$

initial temperature, $T_1=247+273=520\ K$

polytropic index, $n=1.2$

initial volume, $V_1=0.1\ m^3$

work done in the process, $W=50000\ J$

For heat interaction during the polytropic process we have:

$Q=W[(\gamma -n)/(\gamma-1) ]$

$Q=50*[(1.4-1.2)/(1.4-1) ]$

$Q=25\ kJ$

For ideal gas we have the Gas Law:

$P_1.V_1=m.R.T_1$

$6* 10^5* 0.1=m.R* 520$

$m.R=115.385\ J.K^(-1)$

For work we have the relation:

$W=m.R.((T_1-T_2))/((n-1))$

putting respective values

$50000=115.385* ((520-T_2))/((1.2-1))$

$T_2=433.33\ K$

We know entropy change:

$\Delta s=(dQ)/(dT)$

$\Delta s=(25)/(520-433.33)$

$\Delta s= 0.2885 J.K^(-1)$

Answer 2

Answer:

Final answer:

The given question involves the thermodynamic process of an internally reversible process of nitrogen gas (N2) at specific pressure and temperature with a constant value of pν1.2. However, the question does not provide enough information to calculate the heat transfer and entropy change accurately.

Explanation:

The given question involves the thermodynamic process of an internally reversible process of nitrogen gas (N2) at specific pressure and temperature with a constant value of pν1.2. In order to determine the heat transfer and entropy change, we need to use the first and second laws of thermodynamics. However, the question does not provide enough information to calculate the heat transfer and entropy change accurately.

Learn more about thermodynamic process here:

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A 0.700-kg particle has a speed of 1.90 m/s at point circled A and kinetic energy of 7.20 J at point circled B. (a) What is its kinetic energy at circled A? 1.2635 Correct: Your answer is correct. J (b) What is its speed at circled B? 4.54 Correct: Your answer is correct. m/s (c) What is the net work done on the particle by external forces as it moves from circled A to circled B?

Answers

Answer:

a). $E_(kA)=1.2635 J$

b). $V_(B)=4.535(m)/(s)$

c). Δ $E_(t)=8.4635 J$

Explanation:

ΔE=kinetic energy

a).

$E_(kA)=(1)/(2)*m*v_(A) ^(2) \n v_(A)=1.9 (m)/(s)\n m=0.70kg\nE_(kA)=(1)/(2)*0.70kg*(1.9 (m)/(s))^(2) \nE_(kA)=1.2635 J$

b).

$E_(kB)=(1)/(2)*m*v_(B) ^(2)$

$V_(B)^(2)=(E_(kB)*2)/(m) \nV_(B)=\sqrt{(E_(kB)*2)/(m)} \nV_(B)=\sqrt{(7.2J*2)/(0.70kg)} \nV_(B)=4.53 (m)/(s)$

c).

net work= EkA+EkB

$E_(t)=E_(kA)+ E_(kB)\nE_(t)=1.2635J+7.2J\nE_(t)=8.4635J$

Telephone signals are often transmitted over long distances by microwaves. What is the frequency of microwave radiation with a wavelength of 3.5 cm?Express your answer in GHz and using two significant figures.
f = ________GHz
Microwave signals are beamed between two mountaintops 52 km apart. How long does it take a signal to travel from one mountaintop to the other?
Express your answer in ms and using two significant figures.
t = ________ms

Answers

Answer:

1) f= 8.6 GHz

2) t= 0.2 ms

Explanation:

Since microwaves are electromagnetic waves, they move at the same speed as the light in vacuum, i.e. 3*10⁸ m/s.
There exists a fixed relationship between the frequency (f) , the wavelength (λ) and the propagation speed in any wave, as follows:

$v = \lambda * f (1)$

Replacing by the givens, and solving for f, we get:

$f =(c)/(\lambda) =(3e8m/s)/(0.035m) = 8.57e9 Hz (2)$

⇒ f = 8.6 Ghz (with two significative figures)

Assuming that the microwaves travel at a constant speed in a straight line (behaving like rays) , we can apply the definition of average velocity, as follows:

$v =(d)/(t) (3)$

where v= c= speed of light in vacuum = 3*10⁸ m/s

d= distance between mountaintops = 52 km = 52*10³ m

Solving for t, we get:

$t = (d)/(c) = (52e3m)/(3e8m/s) = 17.3e-5 sec = 0.173e-3 sec = 0.173 ms (4)$

⇒ t = 0.2 ms (with two significative figures)

7. If the impact of the golf club on the ball in the previous question occurs over a time of 2 x 10 seconds, whatforce does the ball experience to accelerate from rest to 73 m/s?

Answers

Answer:

3.65 x mass

Explanation:

Given parameters:

Time = 20s

Initial velocity = 0m/s

Final velocity = 73m/s

Unknown:

Force the ball experience = ?

Solution:

To solve this problem, we apply the equation from newton's second law of motion:

F = m $(v - u)/(t)$

m is the mass

v is the final velocity

u is the initial velocity

t is the time taken

So;

F = m ( $(73 - 0)/(20)$ ) = 3.65 x mass

Final answer:

To calculate the force experienced by the ball to accelerate from rest to 73 m/s, use Newton's second law of motion.

Explanation:

To calculate the force experienced by the ball to accelerate from rest to 73 m/s, we can use Newton's second law of motion, which states that force equals mass times acceleration (F = m * a).

Since the ball starts from rest, its initial velocity (vi) is 0 m/s. The final velocity (vf) is 73 m/s. The time (t) taken for the impact is given as 2 x 10 seconds. So, the acceleration (a) can be calculated using the formula a = (vf - vi) / t.

Substituting the given values into the equation, we have a = (73 - 0) / (2 x 10) = 3.65 m/s^2.

Now, we can find the force (F) using the formula F = m * a. If the mass of the ball is known, we can substitute it into the equation to find the force experienced by the ball.

Learn more about Force here:

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G Water enters a house through a pipe 2.40 cm in diameter, at an absolute pressure of 4.10 atm. The pipe leading to the second-floor bathroom, 5.20 m above, is 1.20 cm in diameter. The flow speed at the inlet pipe is 4.75 m/s a) What is the algebraic expression for flow speed in the bathroom?
b) Calculate the flow speed in the bathroom.
c) What is algebraic expression for the pressure in the bathroom?
d) Calculate the water pressure in the bathroom. Report your answer in the (atm) unit.

Answers

Answer:

A) A₁ V₁ = A₂V₂

B) V₂ = 19 m /s

C) P₁ + (1/2)ρv₁² = P₂ + (1/2)ρv₂² + (h₂ - h₁ )ρg

D) P₂ = 1.88 atm

Explanation:

A) From the piaget's theory of conservation of volume, we can calculate the rate of flow of water from;

A₁ V₁ = A₂V₂

Where;

A₁ and A₂ are area of cross section V₁ and V₂ are velocity of flow at two places along pipe.

B) Using the formula given in A above, we obtain;

π x 1.2² x 4.75 = π x 0.6² x V₂

V₂ x 0.36 = 6.84

V₂ = 6.84/0.36

V₂ = 19 m /s

c ) To find pressure we shall apply Bernoulli's theorem in fluid dynamics;

P₁ + (1/2)ρv₁² = P₂ + (1/2)ρv₂² + (h₂ - h₁ )ρg

Where;

P₁ and P₂ are pressure at ground and second floor respectively

v₁ and v₂ are velocity at ground and second floor respectively

h₁ and h₂ are height at ground and second floor respectively ρ is density of water.

Thus, plugging in the relevant values to obtain;

4.1 x 10⁵ + (1/2 x 1000 x 4.75²) = P₂ + (1/2 x 1000 x 19²) + (5.2 x 1000 x 9.8)

(4.1 x 10⁵) + (0.11 x 10⁵) = P₂ + (1.8 X 10⁵) + (0.51 X 10

P₂ = 1.9 X 10⁵ N/m² = 1.88 atm

The _______ principle encourages us to resolve a set of stimuli, such as trees across a ridgeline, into smoothly flowing patternsA.) depth perception.
B.) perception.
C.) similarity.
D.) continuity.

Answers

Answer:

Explanation:

Similarity

A piston above a liquid in a closed container has an area of 0.75m^2, and the piston carries a load of 200kg. What will be the external pressure on the upper surface of the liquid?

Answers

Answer:

2613.3 pa

Explanation:

p=F/A

p=ma/A

p=200×9.8/0.75

p=2613.3

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