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3 years ago

Which circuit is a series circuit ?? plz help

Physics

2 answers:

gregori [183]3 years ago

7 0

In a series circuit, all of the components are connected in the same 'loop' and the current only has one direction/path it can flow through.

In the first three options, the current has multiple paths it can go through. So these three circuits are parallel and not series.

In the last option, the current only has one path where it can flow through, so that circuit is in series.

So Circuit D is a series circuit.

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Answer

Circuit D

Ugo [173]3 years ago

5 0

Answer:the last one

Explanation:in a series circuit, only one current flows through it. i.e it connected by a single wire

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As stated in the statement, we will apply energy conservation to solve this problem.

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3 years ago

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2 years ago

Read 2 more answers

A bead slides without friction around a loopthe-loop. The bead is released from a height 21.9 m from the bottom of the loop-the-

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Answer:

Part a)

$v = 12.45 m/s$

Part B)

$F_n = 0.05 N$

Explanation:

Part A)

As we know that the point A lies on the top of the loop

so we will have by energy conservation

$mgH = \frac{1}{2}mv^2 + mg(2R)$

so the speed at point A is given as

$mg(H - 2R) = \frac{1}{2}mv^2$

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Part B)

Now the force equation at point A is given as

$F_n + mg = \frac{mv^2}{R}$

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3 years ago

A flywheel with a diameter of 1.42 m is rotating at an angular speed of 207 rev/min. (a) What is the angular speed of the flywhe

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Answer:

a. 21.68 rad/s b. 30.78 m/s c. 897 rev/min² d. 1085 revolutions

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b. The linear speed of a point on the flywheel is gotten from v = rω where r = radius of flywheel = 1.42 m

So, v = rω = 1.42 m × 21.68 rad/s = 30.78 m/s

c. Using α = (ω₁ - ω)/t where α = angular acceleration of flywheel, ω = initial angular speed of wheel in rev/min = 21.68 rad/s = 207 rev/min, ω₁ = final angular speed of wheel in rev/min = 1410 rev/min = 147.65 rad/s, t = time in minutes = 80.5/60 min = 1.342 min

α = (ω₁ - ω)/t

= (1410 - 207)/(80.5/60)

= 60(1410 - 207)/80.5

= 60(1203)80.5

= 896.65 rev/min² ≅ 897 rev/min²

d. Using θ = ωt + 1/2αt²

where θ = number of revolutions of flywheel. Substituting the values of the variables from above, ω = 207 rev/min, α = 896.65 rev/min² and t = 80.5/60 min = 1.342 min

θ = ωt + 1/2αt²

= 207 × 1.342 + 1/2 × 896.65 × 1.342²

= 277.725 + 807.417

= 1085.14 revolutions ≅ 1085 revolutions

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