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3 years ago

11

Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit

y of 1700 ft/s. For steady-state operation, and neglecting potential energy effects, determine the exit temperature, in F

1 answer:

Ede4ka [16]3 years ago

7 0

Answer:

$386.2^{\circ}F$

Explanation:

We are given that

$P_1=200lbf/in^2$

$P_2=60lbf/in^2$

$v_1=200ft/s$

$v_2=1700ft/s$

$T_1=500^{\circ}F$

$Q=0$

$C_p=1BTU/lb^{\circ}F$

We have to find the exit temperature.

By steady energy flow equation

$h_1+v^2_1+Q=h_2+v^2_2$

$C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}$

$1BTU/lb=25037ft^2/s^2$

Substitute the values

$1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}$

$500+1.598=T_2+115.4$

$T_2=500+1.598-115.4$

$T_2=386.2^{\circ}F$

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An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train. When the engineer first

GarryVolchara [31]

Answer:

The right answer is "1.369 m/s²".

Explanation:

The given values are:

Distance (s)

= 260 m

Initial speed (u)

= 26 m/s

Reaction time (t')

= 0.51 s

During reaction time, the distance travelled by locomotive will be:

⇒ $s'=ut'$

$=26\times 0.51$

$=13.26 \ m$

Remained distance between locomotive and car:

⇒ $x=s-s'$

$=260-13.26$

$=246.74 \ m$

Now,

The final velocity to avoid collection is, V = 0 m/s

From third equation of motion:

⇒ $V^2=u^2+2ax$

On putting the estimated values, we get

⇒ $0=(26)^2+2\times a\times 246.74$

⇒ $0=676+493.48a$

⇒ $493.48a=-676$

⇒ $a=-\frac{676}{493.48}$

⇒ $a=1.369 \ m/s^2$

3 0

3 years ago

A spherical capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has a radi

Svetlanka [38]

Answer: a) 73.41 10^-12 F; b)4.83* 10^3 N/C; c) 3.66 *10^3 N/C

Explanation: To solve this problem we have to consider the following: The Capacity= Charge/Potential Difference

As we know the capacity is value that depend on the geometry of the capacitor, in our case two concentric spheres.

So Potential Difference between the spheres is given by:

ΔV=- $\int\limits^a_b {E} \, dx$

Where E = k*Q/ r^2

so we have $\int\limits^a_b {K+Q1/r} \, dr$

then

Vb-Va=k*Q(1/b-1/a)=kQ (ab/b-a)

Finally using C=Q/ΔV=ab/(k(b-a))

To caclulate the electric firld we first obtain the charge

Q=ΔV*C=120 V*73.41 10^-12 F=8.8 10^-9 C

so E=KQ/r^2 for both values of r

r=12.8 cm ( in meters)

r2=14.7 cm

E(r1)=4.83* 10^3 N/C

E(r2)=3.66 *10^3 N/C

8 0

2 years ago

A 220 kg crate hangs from the end of a rope of length L = 14.0 m. You push horizontally on the crate with a varying force F to m

kifflom [539]

<u>Answer</u>:

(a) magnitude of F = 797 N

(b)the total work done W = 0

(c)work done by the gravitational force = -1.55 kJ

(d)the work done by the pull = 0

(e) work your force F does on the crate = 1.55 kJ

<u>Explanation</u>:

<u>Given</u>:

Mass of the crate, m = 220 kg

Length of the rope, L = 14.0m

Distance, d = 4.00m

<u>(a) What is the magnitude of F when the crate is in this final position</u>

Let us first determine vertical angle as follows

=> $Sin \theta = \frac{d }{L}$

=> $\theta = Sin^{-1} \frac{d}{L}$ =

Now substituting thje values

=> $\theta = Sin^{-1} \frac{4}{12}$ =

=> $\theta = Sin^{-1} \frac{1}{3}$

=> $\theta = Sin^{-1}(0.333)$

=> $\theta = 19.5^{\circ}$

Now the tension in the string resolve into components

The vertical component supports the weight

=> $Tcos\theta = mg$

=> $T = \frac{mg}{cos\theta}$

=> $T = \frac{230 \times 9.8 }{cos(19.5)}$

=> $T = \frac{2254 }{cos(19.5)}$

=> $T = \frac{2254 }{0.9426}$

=>T =2391N

Therefore the horizontal force

$F = TSin(19.5)$

F = 797 N

b) The total work done on it

As there is no change in Kinetic energy

The total work done W = 0

<u>c) The work done by the gravitational force on the crate</u>

The work done by gravity

Wg = Fs.d = - mgh

Wg = - mgL ( 1 - Cosθ )

Substituting the values

= $-230 \times 9.8\times 12 ( 1 - cos(19.5) )$

= $-230 \times 9.8\times 12 ( 1 - 0.9426) )$

= $-230 \times 9.8\times 12 (0.0574)$

= $-230 \times 9.8\times 0.6888$

= $-230 \times 6.750$

= -1552.55 J

The work done by gravity = -1.55 kJ

<u>d) the work done by the pull on the crate from the rope</u>

Since the pull is perpendicular to the direction of motion,

The work done = 0

e)Find the work your force F does on the crate.

Work done by the Force on the crate

WF = - Wg

WF = -(-1.55)

WF = 1.55 kJ

<u>(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)</u>

Here the work done by force is not equal to F*d

and it is equal to product of the cos angle and F*d

So, it is not equal to the product of the horizontal displacement and the answer to (a)

7 0

3 years ago

Which of the following accurately describe some aspect of gravitational waves? Select all that apply.

steposvetlana [31]

<h2>Answers:</h2>

-The first direct detection of gravitational waves came in 2015

-The existence of gravitational waves is predicted by Einstein's general theory of relativity

-Gravitational waves carry energy away from their sources of emission

<h2>Explanation:</h2>

Gravitational waves were discovered (theoretically) by Albert Einstein in 1916 and "observed" for the first time in direct form in 2015 (although the results were published in 2016).

These gravitational waves are fluctuations or disturbances of space-time produced by a massive accelerated body, modifying the distances and the dimensions of objects in an imperceptible way.

In this context, an excellent example is the system of two neutron stars that orbit high speeds, producing a deformation that propagates like a wave,<u> in the same way as when a stone is thrown into the water</u>. So, in this sense, gravitational waves carry energy away from their sources .

Therefore, the correct options are D, E and F.

5 0

3 years ago

You have been assigned to investigate a traffic accident. The masses of car A and car B are 1300 kg and 1200 kg, respectively. C

jarptica [38.1K]

Answer:

The velocity of A before impact = 17.90 m/s

Explanation:

Coefficient of restitution = (speed of seperation)/(speed of approach)

= (v₁ - v₂)/(u₂ - u₁)

where v₁ = velocity of the car A after the impact = ?

v₂ = velocity of the car B after the impact = ?

u₂ = velocity of the car B before the impact = 0 m/s (it was initially at rest)

u₁ = velocity of car A before the impact = ?

First of, we can solve for v₂, the velocity of car B after the impact, from some of the information given in the question.

- Skid marks indicate car B slid 10 m after the impact

- The coefficient of kinetic friction the tires and road is 0.8.

According to the work energy theorem, the work done by frictional force in stopping the car B is equal to the change in kinetic energy of the car B. (All after collision)

W = ΔK.E

ΔK.E = (1/2)(1200)(v₂²) - 0 (final kinetic energy is 0 since the car comes to stop eventually)

ΔK.E = (600v₂²) J

W = F × d

where F = frictional force = μmg = 0.8×1300×9.8 = 10,192 N

d = distance the car skids over before stopping = 10 m

W = 10,192 × 10 = 101,920 J

W = ΔK.E

101,920 = 600v₂²

v₂² = (101920/600) = 169.867

v₂ = 13.03 m/s

But recall,

Coefficient of restitution = (v₁ - v₂)/(u₂ - u₁)

For the sake of convention, we take the direction of car A's initial velocity to be the positive direction.

u₁ = ?

u₂ = 0 m/s

v₁ = ?

v₂ = +13.03 m/s

Coefficient of restitution = 0.4

0.4 = (v₁ - 13.03)/(0 - u₁)

-0.4u₁ = v₁ - 13.03

v₁ = 13.03 - 0.4u₁

But this is a collision. In a collision, the linear momentum is usually conserved.

Momentum before collision = Momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

1300u₁ + (1200×0) = 1300v₁ + (1200×13.03)

1300u₁ + 0 = 1300v₁ + 15639.95

1300u₁ = 1300v₁ + 15639.95

But recall, from the coefficient of restitution relation,

v₁ = 13.03 - 0.4u₁

Substituting this into the momentum balance equation.

1300u₁ = 1300v₁ + 15639.95

1300u₁ = 1300(13.03 - 0.4u₁) + 15639.95

1300u₁ = 16943.28 - 520u₁ + 15639.95

1820u₁ = 32,583.23

u₁ = (32,583.23/1820)

u₁ = 17.90 m/s

Therefore, the velocity of A before impact = 17.90 m/s

Hope this Helps!!!

4 0

3 years ago