23.8 m/s is initial speed, then stops and travels an additional 179 m, what was the magnitude

Two containers (A and B) are in thermal contact with an environment at temperature T = 280 K. The two containers are connected b

zmey [24]

Answer:

The maximum amount of work is $W = 1563.289 \ J$

Explanation:

From the question we are told that

The temperature of the environment is $T = 280\ K$

The volume of container A is $V_A = 2 m^3$

Initially the number of moles is $n = 1.2 \ moles$

The volume of container B is $V_B = 3.5 \ m^3$

At equilibrium of the gas the maximum work that can be done on the turbine is mathematically represented as

$W = P_A V_A ln[ \frac{V_B}{V_A} ]$

Now from the Ideal gas law

$P_A V_A = nRT$

So substituting for $P_A V_A$ in the equation above

$W = nRT ln [\frac{V_B}{V_A} ]$

Where R is the gas constant with a values of $R = 8.314 \ J/mol$

Substituting values we have that

$W = 1.2 * (8.314) * (280) * ln [\frac{3.5}{2} ]$

$W = 1563.289 \ J$

8 0

3 years ago

A car with a mass of 600 kg is traveling at a velocity of 10 m/s. How much kinetic energy does it have?

Viefleur [7K]

Answer:

KE = 30,000 J

Explanation:

KE = $\frac{1}{2}$ mv²

KE = $\frac{1}{2}$ (600)(10)²

KE = $\frac{1}{2}$ (600)(100)

KE = $\frac{1}{2}$ (60000)

KE = 30,000 J

5 0

3 years ago

The valve in the exit pipe is closed . The density of water is 1000kg / m and the gravitational force on free fall of water is 1

Stels [109]

Answer:

300000

Explanation:

p=30x10x1000=30000pascal

7 0

2 years ago

Read 2 more answers

An automobile with an initial speed of 5.13 m/s accelerates uniformly at the rate of 3.0 m/s2. Find the final speed of the car a

Digiron [165]

Answer:

Final velocity v=19.83 m/sec

Explanation:

We have given initial velocity u =5.13 m/sexc

Acceleration of automobile $a=3m/sec^2$

Time t =4.9 sec

We have to find the final velocity v

According to first law of motion v = u+at ,here v is the final velocity , a is acceleration and t is time

So $v=5.13+3\times 4.9=19.83m/sec$

So the final velocity is 19.83 m/sec

7 0

3 years ago

a circular cylinder and isused to maintain a water depth of 4 m. That is, when the water depth exceeds 4 m, thegate opens slight

stiv31 [10]

Answer:

W / A = 39200 kg / m²

Explanation:

For this problem let's use the equilibrium equation of / newton

F = W

Where F is the force of the door and W the weight of water

W = mg

We use the concept of density

ρ = m / V

m = ρ V

The volume of the water column is

V = A h

We replace

W = ρ A h g

On the other side the cylinder cover has a pressure

P = F / A

F = P A

We match the two equations

P A = ρ A h g

P = ρ g h

P = 39200 Pa

The weight of the water column is

W = 1000 9.8 4 A

W / A = 39200 kg / m²

3 0

3 years ago