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Anastaziya [24]
3 years ago
12

A rigid, insulated tank that is initially evacuated is connected though a valve to a supply line that carries steam at 1 MPa and

300∘C. Now the valve is opened, and steam is allowed to flow slowly into the tank until the pressure reaches 1 MPa, at which point the valve is closed.
Determine the final temperature of the steam in the tank, in ∘C.
Physics
1 answer:
meriva3 years ago
6 0

Answer:

Final Temperature of the steam tank = 456.4°C

Explanation:

Assuming it to be a uniform flow process, kinetic and potential energy to be zero, and work done and heat input to be zero also.  We can conclude that,

Enthalpy of the steam in pipe = Internal Energy of the steam in tank

Using the Property tables and Charts - Steam tables,  

At Pressure= 1 MPa and Temperature= 300°C,

Enthalpy = 3051.2 kJ/kg

At Pressure= 1 MPa and Internal Energy= 3051.2 kJ/kg,

Temperature = 456.4°C.

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A vector has an x-component
9966 [12]

Explanation:

The magnitude of a vector v can be found using Pythagorean's theorem.

||v|| = √(vₓ² + vᵧ²)

||v|| = √((-309)² + (187)²)

||v|| ≈ 361

You can find the angle of a vector using trigonometry.

tan θ = vᵧ / vₓ

tan θ = 187 / -309

θ ≈ 149° or θ ≈ 329°

vₓ is negative and vᵧ is positive, so θ must be in the second quadrant.  Therefore, θ ≈ 149°.

4 0
3 years ago
A 65 kg person jumps from a window to a fire net 18 m below, which stretches the net 1.1 m. Assume the net behaves as a simple s
posledela

Answer:

0.03167 m

1.52 m

Explanation:

x = Compression of net

h = Height of jump

g = Acceleration due to gravity = 9.81 m/s²

The potential energy and the kinetic energy of the system is conserved

P_i=P_f+K_s\\\Rightarrow mgh_i=-mgx+\frac{1}{2}kx^2\\\Rightarrow k=2mg\frac{h_i+x}{x^2}\\\Rightarrow k=2\times 65\times 9.81\frac{18+1.1}{1.1^2}\\\Rightarrow k=20130.76\ N/m

The spring constant of the net is 20130.76 N

From Hooke's Law

F=kx\\\Rightarrow x=\frac{F}{k}\\\Rightarrow x=\frac{65\times 9.81}{20130.76}\\\Rightarrow x=0.03167\ m

The net would strech 0.03167 m

If h = 35 m

From energy conservation

65\times 9.81\times (35+x)=\frac{1}{2}20130.76x^2\\\Rightarrow 10065.38x^2=637.65(35+x)\\\Rightarrow 35+x=15.785x^2\\\Rightarrow 15.785x^2-x-35=0\\\Rightarrow x^2-\frac{200x}{3157}-\frac{1000}{451}=0

Solving the above equation we get

x=\frac{-\left(-\frac{200}{3157}\right)+\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}, \frac{-\left(-\frac{200}{3157}\right)-\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}\\\Rightarrow x=1.52, -1.45

The compression of the net is 1.52 m

4 0
3 years ago
A single point charge is placed at the center of an imaginary cube that has 10 cm long edges. The electric flux out of one of th
12345 [234]

Answer:

Explanation:

<u>Given:</u>

Length of each side of the cube, L=10\ cm

The Elecric flux through one of the side of the cube is, \phi =-1 kNm^2/C.

The net flux through a closed surface is defined as the total charge that lie inside the closed surface divided by \epsilon_0

Since Flux is a scalar quantity. It can added to get total flux through the surface.

\phi_{total}=\dfrac{Q_{in}}{\epsilon_0}\\6\times {-1}\times 10^3=\dfrac{Q_{in}}{\epsilon_0}\\\\Q_{in}=-6}\times 10^3\epsilon_0\\Q_{in}=-6}\times 10^3\times8.85\times 10^{-12}\\Q_{in}=-53.1\times10^{-9}\ C

So the the charge at the centre is calculated.

5 0
3 years ago
Directions: Complete the concept map using the terms listed below.
Komok [63]

Answer:

this is confusing

Explanation:

6 0
3 years ago
Atoms are very small. Which measurement is the approximate diameter of a helium atom?
stich3 [128]
Atoms diameter are on the order of 60 to 600pm  (picometers).  This is around 6*10^-11.
3 0
3 years ago
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