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salantis [7]
2 years ago
3

Two kilograms of nitrogen (N2) at 25°C is contained in a 0.62 m3 rigid tank. This tank is connected by a valve to a 0.16 m3 rigi

d tank containing 0.8 kg of oxygen (O2) at 127°C. The valve is opened, and the gases are allowed to mix, achieving an equilibrium state at 87°C.
initial pressures of N2 is 5.7293 bar and O2 is 5.2 bar.
the final pressure is 6.44 bar.
the magnitude of the heat transfer for the process is 162.8 kJ, and the direction of energy flow is going in.
What is the entropy change for the mixing process, in kJ/K?
Physics
1 answer:
nydimaria [60]2 years ago
3 0

Answer:

Explanation:

For entropy change the formula is

ΔS = ΔQ / T

ΔQ = Δ H

ΔS = Δ H / T

Given

Δ H = + 162.8 kJ

We can take equilibrium temperature as average temperature of the whole process

So, T = 273 + 87 = 360 K

ΔS = Δ H / T

=  162.8 kJ  / 360

= +  0.508 kJ / K .

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Step 2: calculate A and B magnitudes
Step 3: calculate x, y components
Step 4: sum vector components
Step 5: calculate magnitude of R
Step 6: calculate direction of R
4 0
2 years ago
Read 2 more answers
For a magnetic field strength of 2T, estimate the magnitude of the maximum force on a 1-mm-long segment of a single cylindrical
nydimaria [60]

Answer:

Incomplete question

This is the complete question

For a magnetic field strength of 2 T, estimate the magnitude of the maximum force on a 1-mm-long segment of a single cylindrical nerve that has a diameter of 1.5 mm. Assume that the entire nerve carries a current due to an applied voltage of 100 mV (that of a typical action potential). The resistivity of the nerve is 0.6ohms meter

Explanation:

Given the magnetic field

B=2T

Lenght of rod is 1mm

L=1/1000=0.001m

Diameter of rod=1.5mm

d=1.5/1000=0.0015m

Radius is given as

r=d/2=0.0015/2

r=0.00075m

Area of the circle is πr²

A=π×0.00075²

A=1.77×10^-6m²

Given that the voltage applied is 100mV

V=0.1V

Given that resistive is 0.6 Ωm

We can calculate the resistance of the cylinder by using

R= ρl/A

R=0.6×0.001/1.77×10^-6

R=339.4Ω

Then the current can be calculated, using ohms law

V=iR

i=V/R

i=0.1/339.4

i=2.95×10^-4 A

i=29.5 mA

The force in a magnetic field of a wire is given as

B=μoI/2πR

Where

μo is a constant and its value is

μo=4π×10^-7 Tm/A

Then,

B=4π×10^-7×2.95×10^-4/(2π×0.00075)

B=8.43×10^-8 T

Then, the force is given as

F=iLB

Since B=2T

F=iL(2B)

F=2.95×10^-4×2×8.34×10^-8

F=4.97×10^-11N

7 0
2 years ago
A simple watermelon launcher is designed as a spring with a light platform for the watermelon. When an 8.00 kg watermelon is put
olga_2 [115]

To solve this problem it is necessary to apply the concepts related to the Force from Hooke's law, the force since Newton's second law and the potential elastic energy.

Since the forces are balanced the Spring force is equal to the force of the weight that is

F_s = F_g

kx = mg

Where,

k = Spring constant

x = Displacement

m = Mass

g = Gravitational Acceleration

Re-arrange to find the spring constant

k = \frac{mg}{x}

k = \frac{8*9.8}{0.1}

k = 784N/m

Just before launch the compression is 40cm, then from Potential Elastic Energy definition

PE = \frac{1}{2} kx^2

PE =\frac{1}{2} 784*0.4^2

PE = 63.72J

Therefore the energy stored in the spring is 63.72J

6 0
2 years ago
A man pushes his child in a grocery cart. The total mass of the cart and child is 30.0 kg. If the force resisting the carts moti
-BARSIC- [3]

The force applied by the man is 60 N

Explanation:

We can solve this problem by applying Newton's second law, which states that:

\sum F = ma (1)

where

\sum F is the net force acting on the child+cart

m is the mass of the child+cart system

a is their acceleration

In this problem, we have:

m = 30.0 kg is the mass

a=1.50 m/s^2

And there are two forces acting on the child+cart system:

  • The forward force of pushing, F
  • The force resisting the cart motion, R = 15.0 N

Therefore we can write the net force as

\sum F = F -R

where R is negative since its direction is opposite to the motion

So eq.(1) can be rewritten as

F-R=ma

And solving for F,

F=ma+R=(30.0)(1.50)+15.0=60 N

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

4 0
2 years ago
a partially inflated weather balloon has a volume of 1.56 * 10^3 L and a pressure of 98.9 lPa. What is the volume of the balloon
e-lub [12.9K]
To solve the problem, we enumerate all the given first. Then the required and lastly the solution.

Given:

V1= 1.56x10^3 L = 1560 L                P2 = 44.1 kPa
P1 = 98.9 kPa

Required:  V2

Solution:

Assuming the gas is ideal. Ideal gas follows Boyle's Law which states that at a given temperature the product of pressure and volume of a gas is constant. In equation,

PV = k

Applying to the problem, we have
 
P1*V1 = P2*V2
(98.9 kPa)*(1560 L) = (44.1 kPa)*V2
V2 = 3498.5 L

<em>ANSWER: V2 = 3498.5 L</em>
7 0
2 years ago
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