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2 years ago

Does the u declined adverb express time, place, or manner? The squirll spotted us and ran away.

1 answer:

3 0

I feel like it expresses time, but if there is another answer and they say otherwise, then they are probably right.
Hope this helps!

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Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500°C, and 80 m/s and the exit c

FrozenT [24]

Answer:

a) $\Delta \dot K = 24.570\,kW$ , b) $\dot W_{out} = 12729.15\,kW$ , c) $A_{in} = 0.0136\,m^{2}$

Explanation:

A turbine is a device which works usually in steady state and assumption of being adiabatic means no heat interactions between steam through turbine and surroudings and produce mechanical work from fluid energy. Changes in gravitational energy can be neglected. This system can be modelled after the First Law of Thermodynamics:

$-\dot W_{out} + \dot m \cdot (h_{in} - h_{out}) + \frac{1}{2}\cdot \dot m \cdot (v_{in}^{2}-v_{out}^{2}) = 0$

a) Change in kinetic energy

$\Delta \dot K = \frac{1}{2}\cdot \dot m \cdot (v_{in}^{2} - v_{out}^{2})$

$\Delta \dot K = \frac{1}{2} \cdot \left(12.6\,\frac{kg}{s} \right) \cdot \left[\left(80\,\frac{m}{s} \right)^{2}-\left(50\,\frac{m}{s} \right)^{2}\right]$

$\Delta \dot K = 24570\,W$

$\Delta \dot K = 24.570\,kW$

b) Power output

$\dot W_{out} = \dot m \cdot (h_{in} - h_{out}) + \frac{1}{2}\cdot \dot m \cdot (v_{in}^{2}-v_{out}^{2})$

$\dot W_{out} = \left(12.6\,\frac{kg}{s}\right)\cdot \left(3446\,\frac{kJ}{kg} - 2437.7\,\frac{kJ}{kg} \right) + 24.570\,kW$

$\dot W_{out} = 12729.15\,kW$

c) Turbine inlet area

Turbine inlet area can be found by using the following expressions:

$\dot V_{in} = \dot m \cdot \nu_{in}$

$\dot V_{in} = \left(12.6\,\frac{kg}{s}\right) \cdot \left(0.086442\,\frac{m^{3}}{kg} \right)$

$\dot V_{in} = 1.089\,\frac{m^{3}}{s}$

$A_{in} = \frac{\dot V_{in}}{v_{in}}$

$A_{in} = \frac{1.089\,\frac{m^{3}}{s} }{80\,\frac{m}{s} }$

$A_{in} = 0.0136\,m^{2}$

8 0

3 years ago

Ashley decides to enter her pet turtle in a race. She knows her turtle can travel at a rate of 2 meters per hour. The race track

Genrish500 [490]

It would probably be B

7 0

3 years ago

There is a parallel plate capacitor. Both plates are 4x2 cm and are 10 cm apart. The top plate has surface charge density of 10C

liberstina [14]

Answer:

1) The total charge of the top plate is 0.008 C

b) The total charge of the bottom plate is -0.008 C

2) The electric field at the point exactly midway between the plates is 0

3) The electric field between plates is approximately 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N

Explanation:

The given parameters of the parallel plate capacitor are;

The dimensions of the plates = 4 × 2 cm

The distance between the plates = 10 cm

The surface charge density of the top plate, σ₁ = 10 C/m²

The surface charge density of the bottom plate, σ₂ = -10 C/m²

The surface area, A = 0.04 m × 0.02 m = 0.0008 m²

1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C

b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C

2) The electrical field at the point exactly midway between the plates is given as follows;

$V_{tot} = V_{q1} + V_{q2}$

$V_q = \dfrac{k \cdot q}{r}$

Therefore, we have;

The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m

$V_{tot} = \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05} = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0$

The electric field at the point exactly midway between the plates, $V_{tot}$ = 0

3) The electric field, 'E', between plates is given as follows;

$E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C$

E ≈ 1.1294 × 10¹² N/C

The electric field between plates, E ≈ 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates

The charge on an electron, e = -1.6 × 10⁻¹⁹ C

The force on an electron in the middle of the two plates, $F_e$ = E × e

∴ $F_e$ = 1.1294 × 10¹² N/C × -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N

The force on an electron in the middle of the two plates, $F_e$ ≈ 1.807 × 10⁻⁷ N

4 0

2 years ago

A wheel has a radius of 5.9 m. How far (path length) does a point on the circumference travel if the wheel is rotated through an

posledela

Answer:

(a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

Explanation:

Given that,

Radius = 5.9 m

(a). Angle $\theta=30°$

We need to calculate the angle in radian

$\theta=30\times\dfrac{\pi}{180}$

$\theta=0.523\ rad$

We need to calculate the path length

Using formula of path length

$Path\ length =angle\times radius$

$Path\ length=0.523\times5.9$

$Path\ length =3.09\ m$

(b). Angle = 30 rad

We need to calculate the path length

$Path\ length=30\times5.9$

$Path\ length=177\ m$

(c). Angle = 30 rev

We need to calculate the angle in rad

$\theta=30\times2\pi$

$\theta=188.4\ rad$

We need to calculate the path length

$Path\ length=188.4\times5.9$

$Path\ length =1111.56\ m$

Hence, (a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

8 0

2 years ago

Using monochromatic light of 410 nm wavelength, a single thin slit forms a diffraction pattern, with the first minimum at an ang

VLD [36.1K]

Answer:

c. 307 nm

Explanation:

angular position of first dark fringe = λ / d , λ is wavelength and d is width of slit .

(40 x π ) / 180 = 410 / d

angular position of second dark fringe = 2 x λ / d , λ is wavelength and d is width of slit .

(60 x π ) / 180 = 2 x λ / d

Dividing these equations

60 / 40 = 2 x λ / 410

λ = 307.5 nm.

5 0

3 years ago