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Elden [556K]
3 years ago
10

A gas undergoes a process in a piston–cylinder assembly during which the pressure-specific volume relation is pv1.1 = constant.

The mass of the gas is 0.4 lb and the following data are known: p1 = 160 lbf/in.2, V1 = 1 ft3, and p2 = 300 lbf/in.2 During the process, heat transfer from the gas is 2.1 Btu. Kinetic and potential energy effects are negligible. Determine the change in specific internal energy of the gas, in Btu/lb.
Physics
1 answer:
Sati [7]3 years ago
8 0

Answer:

\Delta u = 53.99 Btu/lbm

Explanation:

given data:

P_1 = 160 lbf/in^2 = 23040 lbf/ft^2

P_2 = 300 lbf/in^2 = 56160 lbf/ft^2

V_1 = 1ft^3

Q = - 2.1 Btu

PV^{1.1} = constant

P_1V_1^{1.1} = P_2V_2^{1.1}

V_2 = V_1 [\frac{P_1}{P_2}]^{1.1}

       =  1 [\frac{23040}{56160}]^{\frac{1}{1.1}

       = 0.44 ft^3

work done during polytropic process

w = \frac{P_2V_2 - P_1V_1}{1-n}

   = \frac{56160 \times 0.4759 - 23040\times 1}{1 - 1.2}

   = -18442.1 ft lbf

we know 1 Btu = 778.169 ft. lbf, therefore

w = -23.6993 Btu

Q = W + m\Delta u

-2.1 = - 23.6993 + 0.4 \Delta u

\Delta u = 53.99 Btu/lbm

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A hollow metal sphere has 7 cm and 11 cm inner and outer radii, respectively. The surface charge density on the inside surface i
Alexus [3.1K]

Answer:

1)  E = 0 , 2) zero, 3)    E = - 2,162 10⁴ N/C , 4) directed towards the center of the sphere , 5) E = 1.412 10⁴ N / C , 6)direction coming out of the sphere

Explanation:

The electric field is a vector quantity, therefore we can calculate the field due to each charge distribution and add vector  

          E = E₁ + E₂

To calculate each field we can use Gauss's law, which states that the flow is equal to the charge  by the Gaussian surface divided by ε₀

For this case, let's take a sphere as a Gaussian surface

          Ф = ∫E dA = q_{int} /ε₀

 The area of ​​a sphere is

        A = 4π r²

         E = 1 / 4πε₀   q_{int} / r²

1) r = 4 cm

This radius is smaller than the radius of the sphere, therefore the charge inside is zero and therefore the field is zero

            E = 0

2) there is no field

3) r = 8 cm

Let's calculate each field, for the inner surface

This radius is larger than the internal radius, so the field is

           σ = q_{int} / A

The area of ​​the sphere is

          V = 4 π R_in²

         Rho = q_{int} / 4π R_in²

          q_{int} = ρ 4π R_in²

         E₁ = 1 /ε₀ ρ r_in² / r²

For the outer surface

This radius is smaller so there is no load inside the Gaussian surface and therefore the field is zero

          E₂ = 0

Total E

         E = E₁ + 0

          E = 1 /ε₀  ρ₁ R_in² / r²

Let's calculate

           E = 1 /8,854 10⁻¹²   250 10⁻⁹ (7/8)²

           E = - 2,162 10⁴ N / C

4) as the electric field is negative, it is directed towards the center of the sphere

5) r = 12 cm

In this case the two surfaces contribute to the electric field,

Inner surface

        Q₁ = ρ₁ 4π R_in²

        E₁ = 1 /ε₀  ₁rho1 R_in² / r²

Outer surface

         Q₂ = ρ₂ 4π R_out²

          E₂ = 1 /ε₀  ρ₂ R_out² / r²

The total field is

          E = E₁ + E₂

          E = 1 /ε₀  | ρ | [- R_in² + R_out²] / r²

Let's calculate

          E = 1 /8,854 10⁻¹² 250 10⁻⁹ [- 7² + 11²] / 12²

          E = 1.412 10⁴ N / C

6) As the field is positive, it is directed radially with direction coming out of the sphere

6 0
3 years ago
The risk of earthquakes is high along the Pacific coast of the United States because a. There have been no earthquakes there lat
Ludmilka [50]
(D) That's where the Pacific and North American plates meet.
5 0
3 years ago
How to find a planet’s gravitational field strength using its radius?
grin007 [14]

The gravitational field strength is approximately equal to 10 N.

<u>Explanation:</u>

Gravitational field strength is the measure of gravitational force acting on any object placed on the surface of the planet. Generally, the mass of the object is considered as 1 kg.

So the gravitational field strength will be equal to the gravitational force acting on the object.

The formula for gravitational field strength is

g = \frac{F}{m}

Here g is the gravitational field strength, m is the mass of the object placed on the surface and F is the gravitational force acting on the object.

Since, the mass of any object placed on the surface of earth will be negligible compared to the mass of Earth, so the mass of the object is considered as 1 kg.

Then the g = F

And F =\frac{GMm}{r^{2} }

Here G is the gravitational constant, M is the mass of Earth and m is the mass of the object placed on the surface, while r is the radius of the Earth.

g = F = \frac{6 \times 10^{24} \times 6.67 \times 10^{-11}  \times 1}{(6.6 \times 10^{6}) ^{2} }

g = 0.977 \times 10^1= 9.77\ N

So, the gravitational field strength is approximately equal to 10 N.

5 0
3 years ago
An electron moves at 0.130 c as shown in the figure (Figure 1). There are points: A, B, C, and D 2.10 μm from the electron.
Olegator [25]

Hi there!

We can use Biot-Savart's Law for a moving particle:
B= \frac{\mu_0 }{4\pi}\frac{q\vec{v}\times \vec{r}}{r^2 }

B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)

q = charge of particle (1.6 × 10⁻¹⁹ C)

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

r = distance from particle (2.10 μm)

There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:

B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }

Where 'θ' is the angle between the velocity and radius vectors.

a)
To find the angle between the velocity and radius vector, we find the complementary angle:

θ = 90° - 60° = 30°

Plugging 'θ' into the equation along with our other values:

B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }\\\\B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(30)}{(2.1*10^{-5})^2 }

B = \boxed{7.07 *10^{-10} T}

b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.

c)

In this instance, the radius vector and the velocity vector are perpendicular so

'θ' = 90°.

B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(90)}{(2.1*10^{-5})^2 } = \boxed{1.415 * 10^{-9}T}

d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.

Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.

\boxed{B = 0 T}

3 0
2 years ago
Please help on this one?
dezoksy [38]

the answer is heat engine

3 0
3 years ago
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