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Olin [163]
3 years ago
7

A piston-cylinder device contains 3 kg of saturated water vapor at 200 kPa. Now heat is transferred to the system at constant pr

essure until the temperature reaches 200oC. Calculate the work done by the steam during this process.
Physics
1 answer:
-BARSIC- [3]3 years ago
6 0

Answer:

work done = 117 kJ

Explanation:

given data

mass m = 3 kg

constant pressure P = 200 kPa

temperature T = 200°C

solution

we know that work done by steam is express as

work done = pressure × ΔVolume  ....................1

and here ΔVolume  = final volume - initial volume

we use here steam table and get at pressure 200 kPa

final specific volume = 1.08052 m³/kg

and

initial specific volume = 0.885735 m³/kg

so here

ΔV  = 3 × (1.08052 - 0.885735)

ΔV  = 0.584 m³

so put value in equation 1 we get

work done by steam = 200 × 0.584

work done = 117 kJ

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A centrifuge in a medical research laboratory rotates at an angular speed of 3,650 rev/min. When switched off, it rotates 46.0 t
erma4kov [3.2K]

Answer:

The constant angular acceleration of the centrifuge = -252.84 rad/s²

Explanation:

We will be using the equations of motion for this calculation.

Although, the parameters of this equation of motion will be composed of the angular form of the normal parameters.

First of, we write the given parameters.

w₀ = initial angular velocity = 2πf₀

f₀ = 3650 rev/min = (3650/60) rev/s = 60.83 rev/s

w₀ = 2πf₀ = 2π × 60.83 = 382.38 rad/s

θ = 46 revs = 46 × 2π = 289.14 rad

w = final angular velocity = 0 rad/s (since the centrifuge come rest at the end)

α = ?

Just like v² = u² + 2ay

w² = w₀² + 2αθ

0 = 382.38² + [2α × (289.14)]

578.29α = -146,214.4644

α = (-146,214.4644/578.29)

α = - 252.84 rad/s²

Hope this Helps!!!

6 0
3 years ago
Read 2 more answers
When the forces acting on a particle are resolved into cylindrical components, friction forces always act in the
Aleksandr [31]

Answer:

Tangential

Explanation: This is a kind of force which act on a moving body in such a way that it is curved in the direction of the path of the body. This implies that when the velocity of the object is positive, the acceleration will be negative.

7 0
3 years ago
a man hikes 6.6 km north along a straight path with an average velocity of 4.2 km/h to the north. he rest at a bench for 15 min.
SSSSS [86.1K]

Answer:

2.6h

Explanation:

I attached the image below of the work hope you can see it. Hope this helps!

6 0
2 years ago
a student balances a 1.5kg of broom by placing her finger 1.4m from the end of the broom handle. How far from the broom handle w
Delicious77 [7]

The broom handle that she have to balance if she hung a 400g mass from the end of the broom handle is 5.24m

This problem is centered on moment. Moment is the turning effect of a force about a point. It is expressed as:

Moment = Force× Distance

According to principle of moment, the sum of clockwise moment is equal to sum of anticlockwise moment at shown

M1d1 = M2d2

Given the following

M1 = 1.5kg

d1 = 1.4m

M2 = 400g = 0.4kg

d2 is required

Substitute

1.5(1.4) = 0.4d2

2.1 = 0.4d2

d2 = 2.1/0.4

d2 = 5.24m

Hence the broom handle that she have to  if she hung a 400g mass from the end of the broom handle is 5.24m

Learn more here: brainly.com/question/21945515

4 0
2 years ago
Consult interactive solution 2.22 before beginning this problem. a car is traveling along a straight road at a velocity of +30.0
Inessa05 [86]

Let a_1 be the average acceleration over the first 2.46 seconds, and a_2 the average acceleration over the next 6.79 seconds.

At the start, the car has velocity 30.0 m/s, and at the end of the total 9.25 second interval it has velocity 15.2 m/s. Let v be the velocity of the car after the first 2.46 seconds.

By definition of average acceleration, we have

a_1=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

a_2=\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}

and we're also told that

\dfrac{a_1}{a_2}=1.66

(or possibly the other way around; I'll consider that case later). We can solve for a_1 in the ratio equation and substitute it into the first average acceleration equation, and in turn we end up with an equation independent of the accelerations:

1.66a_2=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

\implies1.66\left(\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}\right)=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

Now we can solve for v. We find that

v=20.8\,\dfrac{\mathrm m}{\mathrm s}

In the case that the ratio of accelerations is actually

\dfrac{a_2}{a_1}=1.66

we would instead have

\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}=1.66\left(\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}\right)

in which case we would get a velocity of

v=24.4\,\dfrac{\mathrm m}{\mathrm s}

6 0
3 years ago
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