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gregori [183]
2 years ago
14

A circuit contains a 40 ohm resistor and a 60 ohms resistor connected in parallel. If you test this circuit with a DMM you shoul

d read a total resistance
Engineering
1 answer:
lorasvet [3.4K]2 years ago
4 0
<h2>Answer:</h2>

24Ω

<h2>Explanation:</h2>

When resistors are connected in parallel, the reciprocal of their combined resistance, when read with a DMM (Digital Multimeter - for measuring various properties of a circuit or circuit element such as resistance...) is the sum of the reciprocals of their individual resistances.

For example if two resistors of resistances R₁ and R₂ are connected together in parallel, the reciprocal of their combined resistance Rₓ is given by;

\frac{1}{R_x} = \frac{1}{R_1} + \frac{1}{R_2}

Solving for Rₓ gives;

R_{x} = \frac{R_1 * R_2}{R_1 + R_2}          ------------------(i)

From the question;

Let

R₁ = resistance of first resistor = 40Ω

R₂ =  resistance of second resistor = 60Ω

Now,

To get their combined or total resistance, Rₓ, substitute these values into equation (i) as follows;  

R_{x} = \frac{40 * 60}{40 + 60}

R_{x} = \frac{2400}{100}

R_{x} = 24 Ω

Therefore, the total resistance is 24Ω

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poizon [28]

Answer:

i) The torque required to raise the load is 15.85 N*m

ii) The torque required to lower the load is 6.91 N*m

iii) The minimum coefficient of friction is -0.016

Explanation:

Given:

dm = mean diameter = 0.03 m

p = pitch = 0.004 m

n = number of starts = 1

The lead is:

L = n * p = 1 * 0.004 = 0.004 m

F = load = 7000 N

dc = collar diameter = 0.035 m

u = 0.05

i) The helix angle is:

tan\alpha =\frac{L}{\pi *d_{m} } =\frac{0.004}{\pi *0.03} \\\alpha =2.43

The torque is:

T=F\frac{d_{m} }{2} (\frac{\pi *u*d_{m}+L }{\pi *d_{m}-uL } )+(u_{c} F+\frac{d_{2} }{2} )=7000*\frac{0.03}{2} (\frac{\pi *0.05*0.03+0.004}{\pi *0.03-0.05*0.004} )+(0.05*7000*\frac{0.035}{2} )=15.85Nm

ii) The torque to lowering the load is:

T=7000*\frac{0.03}{2} (\frac{\pi *0.05*0.03-0.004}{\pi *0.03+0.05*0.004} )+(0.05*7000*\frac{0.035}{2} )=6.91Nm

iii)

T=F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )+u_{c} *F*\frac{d_{c}}{2}\\  0=F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )+u_{c} *F*\frac{d_{c}}{2}\\F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )=-u_{c} *F*\frac{d_{c}}{2}\\\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )=-u_{c} *\frac{d_{c}}{2}\\\\\frac{0.03}{2} (\frac{u*\pi *0.03-0.004}{\pi *0.03+u*0.004} )=-0.05*\frac{0.035}{2}

Clearing u:

u = -0.016

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3 years ago
Architecture reflects multidisciplinary
vovangra [49]

Answer:

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Explanation:

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6 0
2 years ago
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Ede4ka [16]

Answer:58.28 N

Explanation:

Given data

dia. of nozzle \left ( d\right )=2 cm

initial velocity\left ( u\right )=15 m/s

height\left ( h\right )=2m

Now velocity of jet at height of 2m

v^2-u^2=2gh

v^2=15^2-2\left ( 9.81\right )\left ( 2\right )

v=\sqrt{185.76}=13.62 m/s

Now\ forces\ on\ plate\ are\ weight\left ( Downward\right ) and jet\ force\left ( upward\right )

equating them

W=\left ( \rho Av\right )v

W=10^{3}\times \frac{\pi}{4}\left ( 0.02\right )^2\times 13.62^2

W=58.28 N

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3 years ago
A rigid insulated tank is divided into 2 equal compartments by a thin rigid partition. One of the compartments contains air, ass
Illusion [34]
Https://www.slader.com/discussion/question/an-insulated-rigid-tank-is-divided-into-two-equal-parts-by-a-partition-initially-one-part-contains-4/



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2 years ago
wo companies, Ajax Co. and Boho Inc., were negotiating a merger. In the course of the negotiations, an Ajax representative told
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Answer:

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