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Phantasy [73]
2 years ago
8

Steam at a pressure of 15 bar and a temperature of 320oC is contained in a large vessel. Connected to the vessel through a valve

is a turbine followed by a small initially evacuated tank with a volume of 0.6 m3. When emergency power is required, the valve is opened and the tank fills with steam until the pressure is 15 bar. The temperature in the tank is then 400oC. The filling process takes place adiabatically and kinetic and potential energy effects are negligible. Determine the amount of work developed by the turbine, in kJ.

Physics
1 answer:
Luda [366]2 years ago
7 0

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

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6 0
2 years ago
Read 2 more answers
A particle moves along a straight line with a velocity in meters per second given by v = 12 - 3t + 8t, where t is in seconds. Wh
elena-14-01-66 [18.8K]

Answer:

Explanation:

Given the velocity of a particle modeled by the equation

v = 13-3t+8t² where t is in seconds

Given t = 0 and position s = +5m

A) To get the position as a function of time, we will integrate the function with respect to t ad shown;

v = 13-3t+8t²

S = ∫13-3t+8t² dt

S = 13t-13t²/2+8t³/3 + C

at t = 0 and S = +5m

5 = 13(0)-13(0)²/2+8(0)³/3+C

5 = 0-0+0+C

C = 5

Substituting c = 5 into the displacement function

S = 13t-13t²/2+8t³/3 + C

S = 13t-13t²/2+8t³/3 + 5

B) acceleration is the change in velocity with respect to time.

a = dv/dt

Given v = 13-3t+8t²

a= dv/dt = -3+16t

a = 16-3t

C) acceleration at t = 6s is derived by plugging in t = 6 into the resulting equations in (B)

a = 16-3t

a = 16-3(6)

a = 16-18

a = -2m/s²

D) net displacement from t = 0 to t = 6s

At t = 0:

S(0) = 13(0)-13(0)²/2+8(0)³/3 + 5

S(0) = 0+5

S(0) = 5m

At t = 6s

S(6) = 13(6)-13(6)²/2+8(6)³/3 + 5

S(6) = 78-234+576+5

S(6) = 425m

Net displacement from t = 0s to t = 6s is s(6)-s(0)

= 425-5

= 420m

E) Total distance travelled D = S(6)+S(0)

= 425+5

= 430m

F) Average velocity = ∆S/∆t

Average velocity = S(6)-S(0)/6-0

Average velocity = 425-5/6

Average velocity = 420/6

Average velocity = 70m/s

4 0
2 years ago
A 1150 kg car is on a 8.70° hill. using x-y axis tilted down the plane, what is the x-component of the weight?
Fed [463]
I assume the x-y axis are tilted such that the x-axis is parallel to the surface of the hill while the y-axis is perpendicular to it.

In this case, the x-component of the weight is given by:
W_x =mg \sin \theta
where
m is the mass of the car
g is the acceleration of gravity
\theta is the angle of the hill

Substituting numbers into the formula, we find
W_x=(1150 kg)(9.81 m/s^2)(\sin 8.70^{\circ})=1706 N
6 0
2 years ago
A car is traveling at 20.0 m/s on tires with a diameter of 70.0 cm. The car slows down to a rest after traveling 300.0 m. If the
cupoosta [38]

Answer: deceleration of 1.904\ rad/s^2

Explanation:

Given

Car is traveling at a speed of u=20 m/s

The diameter of the car is d=70 cm

It slows down to rest in 300 m

If the car rolls without slipping, then it must be experiencing pure rolling i.e. a=\alpha \cdot r

Using the equation of motion

v^2-u^2=2as\\

Insert v=0,u=20,s=300

0-(20)^2=2\times a\times 300\\\\a=\dfrac{-400}{600}\\\\a=-\dfrac{2}{3}\ m/s^2

Write acceleration as a=\alpha \cdot r

-\dfrac{2}{3}=\alpha \times 0.35\\\\\alpha =-\dfrac{2}{1.05}\\\\\alpha =-1.904\ rad/s^2

So, the car must be experiencing the deceleration of 1.904\ rad/s^2.

4 0
2 years ago
A student decides to spend spring break by driving 50 miles due east, then 50 miles 30 degrees south of east, then 50 miles 30 d
stira [4]

Answer:600 miles, 12

Explanation: The movement described in the question exhibits that of a polygon. Exhibiting a constant distance and angle with only varying direction until the starting point is reached.

The sum of exterior angles of a polygon = 360 degrees.

Exterior angle of a polygon = (360 ÷ number of sides)

Therefore,

Number of sides = 360 ÷ exterior angle

Exterior angle = 30 degrees

Hence,

Number of sides = 360 ÷ 30 = 12 sides

Since distance traveled of 50 miles is the same for each displacement ;

Total displacement = distance traveled * number of sides

Total displacement = 50 * 12 = 600 miles.

5 0
2 years ago
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