Which of the following correctly describes caster?

Name five or more items that were and may still be made by blacksmith

IgorLugansk [536]

Tools, weapons, hardware, armor

8 0

3 years ago

Consider the following list. list = {24, 20, 10, 75, 70, 18, 60, 35} Suppose that list is sorted using the insertion sort algori

Greeley [361]

Answer:

Option B

10,20,24,75,70,18,60,35

Explanation:

The first, second and third iteration of the loop will be as follows

insertion sort iteration 1: 20,24,10,75,70,18,60,35

insertion sort iteration 2:10,20,24,75,70,18,60,35

insertion sort iteration 3: 10,20,24,75,70,18,60,35

8 0

3 years ago

Meeeeep

lesya [120]

THANKYOUUU :)

5 0

2 years ago

Read 2 more answers

A steel wire is suspended vertically from its upper end. The wire is 400 ft long and has a diameter of 3/16 in. The unit weight

jek_recluse [69]

Answer:

a) the maximum tensile stress due to the weight of the wire is 1361.23 psi

b) the maximum load P that could be supported at the lower end of the wire is 624.83 lb

Explanation:

Given the data in the question;

Length of wire L = 400 ft = ( 400 × 12 )in = 4800 in

Diameter d = 3/16 in

Unit weight w = 490 pcf

First we determine the area of the wire;

A = π/4 × d²

we substitute

A = π/4 × (3/16)²

A = 0.0276 in²

Next we get the Volume

V = Area × Length of wire

we substitute

V = 0.0276 × 4800

V = 132.48 in³

Weight of the steel wire will be;

W = Unit weight × Volume

we substitute

W = 490 × ( 132.48 / 12³ )

W = 490 × 0.076666

W = 37.57 lb

a) the maximum tensile stress due to the weight of the wire;

σ $_w$ = W / A

we substitute

σ $_w$ = 37.57 / 0.0276

= 1361.23 psi

Therefore, the maximum tensile stress due to the weight of the wire is 1361.23 psi

b) the maximum load P that could be supported at the lower end of the wire. Allowable tensile stress is 24,000 psi

Maximum load P that the wire can safely support its lower end will be;

P = ( σ $_{all$ - σ $_w$ )A

we substitute

P = ( 24000 - 1361.23 )0.0276

P = 22638.77 × 0.0276

P = 624.83 lb

Therefore, the maximum load P that could be supported at the lower end of the wire is 624.83 lb

5 0

2 years ago

A 3-phase induction motor with 4 poles is connected to a voltage source with an amplitude of 209 Vrms and a frequency of 120 Hz.

Ket [755]

Answer:

Explanation:

Calculating Nsync (Synchronous Speed):

$Nsync = 120f/P$

$Nsync = 120 x 120 / 4\\Nsync = 3600 rpm$

$Wsync = 3600 * 2\pi /180\\Wsync = 377 rad/s$

Calculating s (Slip):

$s = (Nsync - Nm) / Nsync \\[tex]s = (3600-2464)/3600\\s = 0.3156$

Calculating Vth (Thevenin Voltage):

$Vth = Vph (Xm / \sqrt{Rs^{2} + (Xs+Xm)^{2} })\\Vth = 209 (76 / \sqrt{(2.3)^{2} + (1.3 + 76)^{2} }\\Vth = 205.39 V$

Calculating Rth (Thevenin Resistance):

$Rth = Rs (Xm/Xs + Xm)^{2} \\Rth = 2.3 (76/1.3 + 76)^{2} \\Rth = 2.22 ohm$

Calculating Xth (Thevenin Reactance):

Xth = Xs = 1.3 ohm

Calculating Torque:

$T = (3Vth^{2}Rr/s) / (Wsync[(Rth+Rr/s)^{2} + (Xth + Xr)^{2}])\\T = (3*205.39*0.7/0.3156) / 377[(2.22+0.7/0.315)^{2} + (1.3+1.8)^{2}]$

T = 280699 / 377 [19.69 + 9.61]

4 0

3 years ago