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2 years ago

Can I get an answer to this question please

Engineering

1 answer:

crimeas [40]2 years ago

7 0

Answer:

(i) 12 V in series with 18 Ω.

(ii) 0.4 A; 1.92 W

(iii) 1,152 J

(iv) 18Ω — maximum power transfer theorem

Explanation:

As seen by the load, the equivalent source impedance is ...

10 Ω + (24 Ω || 12 Ω) = (10 +(24·12)/(24+12)) Ω = 18 Ω

The open-circuit voltage seen by the load is ...

(36 V)(12/(24 +12)) = 12 V

The Thevenin's equivalent source seen by the load is 12 V in series with 18 Ω.

The load current is ...

(12 V)/(18 Ω +12 Ω) = 12/30 A = 0.4 A . . . . load current

The load power is ...

P = I^2·R = (0.4 A)^2·(12 Ω) = 1.92 W . . . . load power

10 minutes is 600 seconds. At the rate of 1.92 J/s, the electrical energy delivered is ...

(600 s)(1.92 J/s) = 1,152 J

The load resistance that will draw maximum power is equal to the source resistance: 18 Ω. This is the conclusion of the Maximum Power Transfer theorem.

The power transferred to 18 Ω is ...

((12 V)/(18 Ω +18 Ω))^2·(18 Ω) = 144/72 W = 2 W

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If block A of the pulley system is moving downward at 6 ft>s while block C is moving down at 18 ft>s, determine the relati

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Answer:

Explanation:

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3 years ago

A number 12 copper wire has a diameter of 2.053 mm. Calculate the resistance of a 37.0 m long piece of such wire.

Alinara [238K]

Answer:

R=1923Ω

Explanation:

Resistivity(R) of copper wire at 20 degrees Celsius is 1.72x10^-8Ωm.

Coil length(L) of the wire=37.0m

Cross-sectional area of the conductor or wire (A) = πr^2

A= π * (2.053/1000)/2=3.31*10^-6

To calculate for the resistance (R):

R=ρ*L/A

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Approximately, R=1923Ω

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3 years ago

Scenario* The organization is a regional XYZ Credit Union/Bank that has multiple branches and locations throughout the region.*

fiasKO [112]

Answer:

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Both would include the OAS statements and strategic shifts.

3) The OAS and strategic shift would help to create a balanced score card that would help for data gathering and analysis

This would include the financial, customers goal, Process goals, people's goals etc.

4) To create and develop the plan's framework:

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3 years ago

Two identical billiard balls can move freely on a horizontal table. Ball a has a velocity V0 and hits balls B, which is at rest,

Lyrx [107]

Answer:

Velocity of ball B after impact is $0.6364v_0$ and ball A is $0.711v_0$

Explanation:

$v_0$ = Initial velocity of ball A

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$v_B$ = Initial velocity of ball B = 0

$(v_A)_n'$ = Final velocity of ball A

$v_B'$ = Final velocity of ball B

$e$ = Coefficient of restitution = 0.8

From the conservation of momentum along the normal we have

$mv_A+mv_B=m(v_A)_n'+mv_B'\\\Rightarrow v_0\cos45^{\circ}+0=(v_A)_n'+v_B'\\\Rightarrow (v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0$

Coefficient of restitution is given by

$e=\dfrac{v_B'-(v_A)_n'}{v_A-v_B}\\\Rightarrow 0.8=\dfrac{v_B'-(v_A)_n'}{v_0\cos45^{\circ}}\\\Rightarrow v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0$

$(v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0$

$v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0$

Adding the above two equations we get

$2v_B'=\dfrac{1.8}{\sqrt{2}}v_0\\\Rightarrow v_B'=\dfrac{0.9}{\sqrt{2}}v_0$

$\boldsymbol{\therefore v_B'=0.6364v_0}$

$(v_A)_n'=\dfrac{1}{\sqrt{2}}v_0-0.6364v_0\\\Rightarrow (v_A)_n'=0.07071v_0$

From the conservation of momentum along the plane of contact we have

$(v_A)_t'=(v_A)_t=v_0\sin45^{\circ}\\\Rightarrow (v_A)_t'=\dfrac{v_0}{\sqrt{2}}$

$v_A'=\sqrt{(v_A)_t'^2+(v_A)_n'^2}\\\Rightarrow v_A'=\sqrt{(\dfrac{v_0}{\sqrt{2}})^2+(0.07071v_0)^2}\\\Rightarrow \boldsymbol{v_A'=0.711v_0}$

Velocity of ball B after impact is $0.6364v_0$ and ball A is $0.711v_0$ .

5 0

3 years ago

A three-point bending test is performed on a glass specimen having a rectangular cross section of height d 5 mm (0.2 in.) and wi

Anon25 [30]

Answer:

The flexural strength of a specimen is = 78.3 M pa

Explanation:

Given data

Height = depth = 5 mm

Width = 10 mm

Length L = 45 mm

Load = 290 N

The flexural strength of a specimen is given by

$\sigma = \frac{3 F L}{2 bd^{2} }$

$\sigma = \frac{3(290)(45)}{2 (10)(5)^{2} }$

$\sigma =$ 78.3 M pa

Therefore the flexural strength of a specimen is = 78.3 M pa

4 0

3 years ago