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Andrej [43]
3 years ago
6

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

d 300 ºC. 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.
Physics
1 answer:
ss7ja [257]3 years ago
3 0

Answer:

Explanation:

The steam was earlier at 300 ° C. and pressure of 1 MPa. When the gas is allowed to expand against vacuum , work done by the gas is nil because there is no external pressure against which it has to work . Therefore there will not be any change in its internal energy. Since the tank is insulated therefore there is no possibility of external heat to increase its internal energy.

Hence the temperature of gas will remain unaffected.  It will remain stagnant at 300

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Fill in the blanks to complete each statement about the heating of Earth's surface.
cricket20 [7]

Answer:

absorption and insolation.

Explanation:

6 0
4 years ago
A 1,70 atm, una muestra de gas ocupa 4,25 litros. Si la presión en el gas aumenta a 2.40 atm, ¿cuál será el nuevo volumen?
statuscvo [17]

At 1.70 atm, a gas sample occupies 4.25 liters. If the pressure in the gas increases to 2.40 atm, what will the new volume be?

Answer:

3.01L

Explanation:

Given parameters:

Initial pressure, P1  = 1.7atm

Initial volume, V1  = 4.25L

Final pressure, P2  = 2.4atm

Unknown:

Final or new volume, V2  = ?

Solution:

To solve this problem, we use Boyle's law which states that "the volume of a fixed mass of a gas varies inversely as the pressure changes, if the temperature is constant".

            P1 V1  = P2 V2

P1 is the initial pressure

V1 is the initial volume

P2 final pressure

V2 final volume

        1.7 x 4.25  = 2.4 x V2

             V2  = 3.01L

8 0
3 years ago
A city of punjab has 15 percemt chance of wet weather on any given day. What is probability that it will take a week for it thre
sesenic [268]

Answer:

The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447

Explanation:

We are given that A city of Punjab has 15 percent chance of wet weather on any given day.

So, Probability of wet weather = 0.15

Probability of not being a wet weather = 1-0.15 =0.85

We are supposed to find probability that it will take a week for it three wet weather on 3 separate days

Total number of days in a week = 7

We will use binomial over here

n = 7

p =probability of failure = 0.15

q = probability of success=0.85

r=3

Formula :P(r=3)=^nC_r p^r q ^{n-r}

P(r=3)=^{7}C_{3} (0.15)^3 (0.85)^{7-3}\\P(r=3)=\frac{7!}{3!(7-3)!} (0.15)^3 (0.85)^{7-3}\\P(r=3)=0.06166

Standard deviation =\sqrt{n \times p \times q}

Standard deviation =\sqrt{7 \times 0.15 \times 0.85}

Standard deviation =0.9447

Hence The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447

4 0
3 years ago
A railroad track and a road cross at right angles. An observer stands on the road and watches an eastbound train traveling at 60
mamaluj [8]

Answer:

After 4 s of passing through the intersection, the train travels with 57.6 m/s

Solution:

As per the question:

Suppose the distance to the south of the crossing watching the east bound train be x = 70 m

Also, the east bound travels as a function of time and can be given as:

y(t) = 60t

Now,

To calculate the speed, z(t) of the train as it passes through the intersection:

Since, the road cross at right angles, thus by Pythagoras theorem:

z(t) = \sqrt{x^{2} + y(t)^{2}}

z(t) = \sqrt{70^{2} + 60t^{2}}

Now, differentiate the above eqn w.r.t 't':

\frac{dz(t)}{dt} = \frac{1}{2}.\frac{1}{sqrt{3600t^{2} + 4900}}\times 2t\times 3600

\frac{dz(t)}{dt} = \frac{1}{sqrt{3600t^{2} + 4900}}\times 3600t

For t = 4 s:

\frac{dz(4)}{dt} = \frac{1}{sqrt{3600\times 4^{2} + 4900}}\times 3600\times 4 = 57.6\ m/s

4 0
3 years ago
Suppose a nonconducting sphere, radius r2, has a spherical cavity of radius r1 centered at the sphere's center. Assuming the cha
leva [86]

Answer:

Explanation:

a ) Between r = 0 and r = r₁

Electric field will be zero . It is so because no charge lies in between r = 0 and r = r₁ .

b ) From r = r₁ to r = r₂

At distance r , charge contained in the sphere of radius r

volume charge density x 4/3 π r³

q = Q x r³ / R³

Applying Gauss's law

4πr² E = q / ε₀

4πr² E = Q x r³ / ε₀R³

E= Q x r / (4πε₀R³)

E ∝ r .

c )

Outside of r = r₂

charge contained in the sphere of radius r = Q

Applying Gauss's law

4πr² E = q / ε₀

4πr² E = Q  / ε₀

E = Q  / 4πε₀r²

E ∝ 1 / r² .

6 0
3 years ago
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