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3 years ago

A. What is the total current of this circuit?

Engineering

1 answer:

Brrunno [24]3 years ago

5 0

A. 0.36 amps
B. 0.36 amps
C. Series Circuit
D. Current always stays the same in a parallel circuit, and you can prove this by dividing total voltage by total resistance (V/R)
E. 15 ohms+10 ohms= 25 ohms

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A rod is 2m long at temperature of 10oC. Find the expansion of the rod, when the temperature is raised to 80oC. If this expansio

Damm [24]

Answer:

ΔL = 1.68 mm

σ = 84 MPa

Explanation:

Thermal expansion is:

ΔL = α ΔT L

Thermal stress is:

σ = α ΔT E

Given:

α = 1.2×10⁻⁵ /°C

E = 1.0×10⁵ MPa

ΔT = 80°C − 10°C = 70°C

L = 2 m

ΔL = (1.2×10⁻⁵ /°C) (70°C) (2 m)

ΔL = 0.00168 m

ΔL = 1.68 mm

σ = (1.2×10⁻⁵ /°C) (70°C) (1.0×10⁵ MPa)

σ = 84 MPa

8 0

4 years ago

What is the difference between the pressure head at the end of a 150m long pipe of diameter 1m coming from the bottom of a reser

uysha [10]

Answer:

$\frac {p_2- p_1}{\rho g} = 31.06 m$

Explanation:

from bernoulli's theorem we have

$\frac{p_1}{\rho g} + \frac{v_1^{2}}{2g} +z_1 = \frac{p_2}{\rho g} + \frac{v_2^{2}}{2g} +z_2 + h_f$

we need to find pressure head difference i.e.

$\frac {p_2- p_1}{\rho g} = (z_1 - z_2) - h_f$

where h_f id head loss

$h_f = \frac{flv^{2}}{D 2g}$

velocity v = $\frac{1}{n} * R^{2/3} S^{2/3}$

$S = \frac{\delta h}{L} = \frac{40}{150} = 0.267$

hydraulic mean radius R = $\frac{A}{P} = \frac{hw}{2h+w}$

$R = \frac{40*1}{2*40+1} = 0.493 m$

so velocity is = $\frac{1}{0.013} * 0.493^{2/3} 0.267^{1/2}$

v = 24.80 m/s

head loss

$h_f = \frac{0.0019*150*24.80^{2}}{1* 2*9.81}$

$h_f =8.93 m$

pressure difference is

$\frac {p_2- p_1}{\rho g} = 40 - 8.93 = 31.06 m$

$\frac {p_2- p_1}{\rho g} = 31.06 m$

4 0

3 years ago

A large well-mixed tank of unknown volume, open to the atmosphere initially, contains pure water. The initial height of the solu

trasher [3.6K]

Answer:

The exact time when the sample was taken is = 0.4167337 hr

Explanation:

The diagram of a sketch of the tank is shown on the first uploaded image

Let A denote the first inlet

Let B denote the second inlet

Let C denote the single outflow from the tank

From the question we are given that the diameter of A is = 1 cm = 0.01 m

Area of A is = $\frac{\pi}{4}(0.01)^{2} m^{2}$

= $7.85 *10^{-5}m^{2}$

Velocity of liquid through A = 0.2 m/s

The rate at which the liquid would flow through the first inlet in terms of volume = $\frac{Volume of Inlet }{time}$ = $Velocity * Area$ i.e is $m^{2} * \frac{m}{s} = \frac{m^{3}}{s}$

= $0.2 *7.85*10^{-5} \frac{m^{3}}{s}$

The rate at which the liquid would flow through the first inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

= $1039.8 * 0.2 * 7.85 *10^{-5} Kg/s$

= 0.016324 $\frac{Kg}{s}$

From the question the diameter of B = 2 cm = 0.02 m

Area of B = $\frac{\pi}{4} * (0.02)^{2} m^{2} = 3.14 * 10^{-4}m^{2}$

Velocity of liquid through B = 0.01 m/s

The rate at which the liquid would flow through the first inlet in terms of volume = $\frac{Volume of Inlet }{time}$ = $Velocity * Area$ i.e is $m^{2} * \frac{m}{s} = \frac{m^{3}}{s}$

= $3.14*10^{-4} *0.01 \frac{m^{3}}{s}$

The rate at which the liquid would flow through the second inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

= $1053 * 3.14*10^{-6} \frac{Kg}{s}$

= 0.00330642 $\frac{Kg}{s}$

From the question The flow rate in term of volume of the outflow at the time of measurement is given as = 0.5 L/s

And also from the question the mass of potassium chloride at the time of measurement is given as 13 g/L

So The rate at which the liquid would flow through the outflow in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

= $13\frac{g}{L} * 0.5 \frac{L}{s}$

= $\frac{6.5}{1000}\frac{Kg}{s}$ Note (1 Kg = 1000 g)

= 0.0065 kg/s

Considering potassium chloride

Let denote the rate at which liquid flows in terms of mass as as $\frac{dm}{dt}$ i.e change in mass with respect to time hence

Input(in terms of mass flow ) - output(in terms of mass flow ) = Accumulation in the Tank(in terms of mass flow )

(0.016324 + 0.00330642) - 0.0065 = $\frac{dm}{dt}$

$\int\limits {\frac{dm}{dt} } \, dx =\int\limits {0.01313122} \, dx$

=> 0.01313122 t = (m - $m_{o}$ )

From the question (m - $m_{o}$ ) is given as = 19.7 Kg

Hence the time when the sample was taken is given as

0.01313122 t = 19.7 Kg

=> t = 1500.2414 sec

t = .4167337 hours (1 hour = 3600 seconds)

5 0

4 years ago

What do you understand by the term phase angle?<br>

aleksandrvk [35]

Answer:

The angle between the earth and the sun as seen from a planet is called phase angle.

5 0

3 years ago

At a certain location, wind is blowing steadily at 5 mph. Suppose that the mass density of air is 0.0796 lbm/ft3 and determine t

nlexa [21]

Answer:

The radius of a wind turbine is 691.1 ft

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

Explanation:

Given;

power generation potential (PGP) = 1000 kW

Wind speed = 5 mph = 2.2352 m/s

Density of air = 0.0796 lbm/ft³ = 1.275 kg/m³

Radius of the wind turbine r = ?

Wind energy per unit mass of air, e = E/m = 0.5 v² = (0.5)(2.2352)²

Wind energy per unit mass of air = 2.517 J/kg

PGP = mass flow rate * energy per unit mass

PGP = ρ*A*V*e

$PGP = \rho *\frac{\pi r^2}{2} *V*e \\\\r^2 = \frac{2*PGP}{\rho*\pi *V*e} , r=\sqrt{ \frac{2*PGP}{\rho*\pi *V*e}} = \sqrt{ \frac{2*10^6}{1.275*\pi *2.235*2.517}}$

r = 210.64 m = 691.1 ft

Thus, the radius of a wind turbine is 691.1 ft

PGP = CVᵃ

For best design of wind turbine Betz limit (c) is taken between (0.35 - 0.45)

Let C = 0.4

PGP = Cvᵃ

take log of both sides

ln(PGP) = a*ln(CV)

a = ln(PGP)/ln(CV)

a = ln(1000)/ln(0.4 *2.2352) = 7.73

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

5 0

3 years ago