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ArbitrLikvidat [17]
4 years ago
7

Five 2.0-ω resistors are connected as shown in the figure. what is the equivalent resistance of this combination between points

a and b?
Physics
1 answer:
Misha Larkins [42]4 years ago
5 0
If resistors are connected in series then total resistance=2+2+2+2+2
=10 watt
if connected in parallel then total resistance,1/R=1/2+1/2+1/2+1/2+1/2
1/R=5/2
So R=2/5
R=0.4w
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If an R = 1-kΩ resistor, a C = 1-μF capacitor, and an L = 0.2-H inductor are connected in series with a V = 150 sin (377t) volts
fgiga [73]

Answer

given,

R = 1-kΩ  = 1000 Ω

C = 1-μF

L = 0.2-H

V = V_max sin( ω t)

comparing

V = 150 sin ( 377 t)

ω = 377

\chi_c = \dfrac{1}{\omega C}

\chi_c = \dfrac{1}{377 \times 1 \times 10^{-6}}

\chi_c = 2652.5\Omega

\chi_L =377 \times 0.2

\chi_L =75.4\ \Omega

Impedance,

Z = \sqrt{R^2+(\chi_L-\chi_c)^2}

Z = \sqrt{1000^2+(75.4 -2652.5)^2}

Z = 2764.3 Ω

now,

V_{max} = 150 V

I_{max} = \dfrac{V}{Z}

I_{max} = \dfrac{150}{2764.3}

I_{max} = 0.0543

I_{max} = 54.3\ mA

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3 years ago
When air is blown across the top of an open
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When air is blown across the top of an open <span>water bottle, air molecules in the bottle vibrate at a particular frequency and sound is produced in a process called "refraction". 
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explain what would be the same and what would be different about the phases and view of Earth to an inhabitant of the near side
guapka [62]

Answer:

Both the Earth and the moon are always half illuminated by the sun. But from either world, at any given time, you can see varying portions of that lighted half – or various phases of the Earth or moon. ... Instead, from a given point on the moon's near side, you'd always see Earth hanging in your sky.

Explanation:

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To practice Problem-Solving Strategy 11.1 Equilibrium of a Rigid Body. A horizontal uniform bar of mass 2.7 kg and length 3.0 m
Zanzabum

Answer:

T_{1} = 14.88 N

Explanation:

Let's begin by listing out the given variables:

M = 2.7 kg, L = 3 m, m = 1.35 kg, d = 0.6 m,

g = 9.8 m/s²

At equilibrium, the sum of all external torque acting on an object equals zero

τ(net) = 0

Taking moment about T_{1} we have:

(M + m) g * 0.5L - T_{2}(L - d) = 0

⇒ T_{2} = [(M + m) g * 0.5L] ÷ (L - d)

T_{2} = [(2.7 + 1.35) * 9.8 * 0.5(3)] ÷ (3 - 0.6)

T_{2}= 59.535 ÷ 2.4

T_{2} = 24.80625 N ≈ 24.81 N

Weight of bar(W) = M * g = 2.7 * 9.8 = 26.46 N

Weight of monkey(w) = m * g = 1.35 * 9.8 = 13.23 N

Using sum of equilibrium in the vertical direction, we have:

T_{1} + T_{2} = W + w   ------- Eqn 1

Substituting T2, W & w into the Eqn 1

T_{1} + 24.81 = 26.46 + 13.23

T_{1} = <u>14.88</u> N

8 0
3 years ago
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