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3 years ago

Find fq, the vertical force that pier q exerts on the right end of the bridge.

1 answer:

5 0

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
View the entire bridge as the system, and find the torque about a pivot chosen to eliminate the torque from one of the unknowns
Find the sum of the torques ∑ τ Q about the right pier Q of the bridge. Remember that counterclockwise torque is positive
∑τQ - 2MgL− 3FpL

The sum of the torques is zero

FP = 2/3 Mg

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Answer:

Explanation:

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3 years ago

The half-wave rectifier circuit of vs(t) = 170 sin(377t) V and a load resistance R = 15Ω. Determine: a. The average load current

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Answer:

a average load current = 11.33 A

b rms load current = 8.02A

c true power =962.64 W

d apparent power =962.64 W

e. power factor cosθ =1

Explanation:

Vs (t) = 170 sin(377t) v

Vm =170v

Vrms = 170/√2 =120.23 v

Im = Vm/R = 170/15 = 11.33 A

Irms = Im/ √2 = 11.33/√2 =8.02A

Resistors are electronic components that consume energy

the power in a resistor is given by P =IVcosθ ; in a resistor cosθ =1

P =IV

The electrical power consumed by a resistance, (R) is called the true or real power

and is obtained by multiplying the rms voltage with the rms current.

P= Vrms × Irms

120.03×8.02

P= 962.64 W ; true power

apparent power = Vrms × Irms

=120.03×8.02

= 962.64W ; apparent power

power factor cosθ = true power/ apparent power

cosθ = 962.64/962.64

cosθ = 1

For the purely resistive circuit, the power factor is 1 , because the reactive power is equal to zero (0).

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A solenoid with an inductance of 8 mH is connected in series with a resistance of 5 Ω and an EMF forming a series RL circuit. A

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Answer:

induced EMF = 240 V

and by the lenz's law direction of induced EMF is opposite to the applied EMF

Explanation:

given data

inductance = 8 mH

resistance = 5 Ω

current = 4.0 A

time t = 0

current grow = 4.0 A to 10.0 A

to find out

value and the direction of the induced EMF

solution

we get here induced EMF of induction is express as

E = - L $\frac{dI}{dt}$ ...................1

so E = - L $\frac{I2 - I1}{dt}$

put here value we get

E = - 8 × $10^{-3}$ $\frac{10 - 4}{0.2*10^{-3}}$

E = -40 × 6

E = -240

take magnitude

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and by the lenz's law we get direction of induced EMF is opposite to the applied EMF

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3 years ago

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Answer:

Hey!

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Explanation:

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