Why do y'all boys be so toxic and fake?

CaHeK987 [17]

Explanation:

An overhead power line is a structure used in electric power transmission and distribution to transmit electrical energy across large distances. It consists of one or more conductors (commonly multiples of three) suspended by towers or poles. Since most of the insulation is provided by air, overhead power lines are generally the lowest-cost method of power transmission for large quantities of electric energy.

<h3>Construction</h3>

Towers for support of the lines are made of wood (as-grown or laminated), steel or aluminum (either lattice structures or tubular poles), concrete, and occasionally reinforced plastics. The bare wire conductors on the line are generally made of aluminum (either plain or reinforced with steel or composite materials such as carbon and glass fiber), though some copper wires are used in medium-voltage distribution and low-voltage connections to customer premises. A major goal of overhead power line design is to maintain adequate clearance between energized conductors and the ground so as to prevent dangerous contact with the line, and to provide reliable support for the conductors, resilience to storms, ice loads, earthquakes and other potential damage causes. Today overhead lines are routinely operated at voltages exceeding 765,000 volts between conductors.

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6 0

4 years ago

When _____ ,the lithium ions are removed from the_____ and added into the _____

bezimeni [28]

Answer:

b. Discharging; anode; cathode

Explanation:

When discharging , it means the battery is producing a flow electric current, the lithium ions are released from the anode to the cathode which generates the flow of electrons from one side to another. When charging Lithium ions are released by the cathode and received by the anode.

8 0

3 years ago

thier only motto and goal is to work for society and not make any profits A.small business entreprenuership B.scalable start up

sweet-ann [11.9K]

Answer:

social entrepreneurship

8 0

3 years ago

Read 2 more answers

A solid circular shaft has a uniform diameter of 5 cm and is 4 m long. At its midpoint 65 hp is delivered to the shaft by means

AlekseyPX

Answer:

A) τ_max = 59.139 x 10^(6) Pa

B) θ = 0.0228 rad.

Explanation:

A) In the left half of the shaft we have 25 hp which corresponds to a torque T1 given by;

P = Tω

Where P is power and ω is angular speed.

Power = 25 HP = 25 x 746 W = 18650W

ω = 200 rev/min = 200 x 0.10472 rad/s = 20.944 rad/s

P = T1•ω

T1 = P/ω = 18650/20.944

T1 = 890.47 N.m

Similarly, in the right half we have 40 hp corresponding to a torque T2

given by;

P = T2•ω

T2 = P/ω

Where P = 40 x 760 = 30,400W

T2 = 30400/20.944 = 1451.49 N.m

The maximum shearing stress consequently occurs in the outer fibers in the right half and is given by;

τ_max = Tρ/J

Where J is polar moment of inertia and has the formula ;J = πd⁴/32

d = 5cm = 0.05m

J = π(0.05)⁴/32 = 6.136 x 10^(-7) m⁴

ρ = 0.05/2 = 0.025m

T will be T2 = 1451.49 N.m

Thus,

τ_max = Tρ/J

τ_max = 1451.49 x 0.025/6.136 x 10^(-7)

τ_max = 59139022.94 N/m² = 59.139 x 10^(6) Pa

B) The angles of twist of the left and right ends relative to the center are, respectively, using θ = TL/GJ

G = 80 Gpa = 80 x 10^(9) Pa

θ1 = (890.47 x 2)/(80 x 10^(9) x 6.136 x 10^(-7)) = 0.0363 rad

Similarly;

θ2 = (1451.49 x 2)/(80 x 10^(9) x 6.136 x 10^(-7)) = 0.0591 rad

Since θ1 and θ2 are in the same direction, the relative angle of twist between the two ends of the shaft is

θ = θ2 – θ1

θ = 0.0591 - 0.0363

θ = 0.0228 rad.

6 0

3 years ago

H. Blasius correlated data on turbulent friction factor in smooth pipes. His equation f s m o o t h ≈ 0.3164 Re − 1 / 4 fsmooth≈

tiny-mole [99]

Answer:

Therefore the angle the pipe needed to make the static pressure constant along the pipe is θ = 4° 16'

Explanation:

The first step to take is to calculate the the velocity of flow through a pipe

Q =Av

Where Q = is the discharge through pipe

A = Area of the pipe

v = the flow of velocity

We substitute 0.001 m^3/s for Q and 0.03 m for D

Q= Av

0.001=Av

Substitute π/4 D² for A

0.001 = π/4 D² (v)

v = 0.004/πD²

D = he diameter of the pipe

substitute 3 cm for D

v= 0.004/π * [3 cm * 1 m/100 cm]²

v =1.414 m/s

Obtain fluid properties from the table Kinematic viscosity and Dynamic of water

p =1000 kg /m³

μ= 1.002 * 10^ ⁻³ N.s/m³

Thus,

we write the expression to determine the Reynolds number of flow

Re = pvD/μ

Re = is the Reynolds number

p =density

μ = dynamic viscosity at 20⁰C

We then substitute 1000 kg /m³ in place of p, 1.002 * 10^ ⁻³ N.s/m³ for μ,

1.414 m/s for v and 0.03 m for D

Thus,

Re = 1000 * 1.414 * 0.03/ 1.002 * 10^ ⁻³ = 42335

The next step is to calculate the friction factor form the Blasius equation

f = 0.3164 (Re)^1/4

f = friction factor

We substitute 42335 for Re

f = 0.3164 (42335)1/4

=0.022

The next step is to write the expression to determine the friction head loss

hl = flv²/2gD

hl = head loss

l = length of pipe

g= acceleration due to gravity

We then again substitute 0.022 for f, 1.414 m/s for v, 0.03 m for D, and 9.8 m/s² for g.

so,

hl = flv²/2gD

hl/L = 0.022 * 1.414²/2 * 9.81 * 0.03

sinθ = 0.07473

θ = 4° 16'

Therefore the angle the pipe needed to make the static pressure constant along the pipe is θ = 4° 16'

3 0

3 years ago