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2 years ago

ANSWER ASAP PLS

Physics

1 answer:

posledela2 years ago

4 0

Answer:

<h3>JAWAB SECEPATNYA pliss</h3><h3 /><h3>Anda memiliki rangkaian paralel 10 volt, dengan 2 resistor di atasnya. Berapakah tegangan pada</h3><h3>resistor pertama? Di seberang kedua?</h3><h3 /><h3>(saya akan menandai tercerdas tolong bantu)</h3>

Explanation:

Hukum Ohm

= tegangan

= kuat arus

= ketahanan

Kalau kamu mau mencari tegangan listrik, kamu gunakan rumus V = I.R. Kalau ternyata kamu perlu mencari kuat arus listrik, maka gunakan rumus I = V/R. Nah, kalau yang kamu cari adalah hambatan listrik, maka gunakan rumus R = V/I.

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Determine the total impedance of an LRC circuit connected to a 10.0- kHz, 725-V (rms) source if L = 36.00 mL, R = 10.00 kÎ©, and

SOVA2 [1]

Answer:

10042.6 ohm

Explanation:

f = 10 kHz = 10000 Hz, L = 36 mH = 0.036 H, R = 10 kilo Ohm = 10000 ohm

C = 5 nF = 5 x 10^-9 F

XL = 2 x π x f x L

XL = 2 x 3.14 x 10000 x 0.036 = 2260.8 ohm

Xc = 1 / ( 2 x π x f x C) = 1 / ( 2 x 3.14 x 10000 x 5 x 10^-9)

Xc = 3184.7 ohm

Total impedance is Z.

Z^2 = R^2 + (XL - Xc)^2

Z^2 = 10000^2 + ( 2260.8 - 3184.7 )^2

Z = 10042.6 ohm

4 0

3 years ago

Determine

djyliett [7]

(i) The total capacitance for the circuit is 5 μF.

(ii) The total charge stored in the circuit is 1 x 10⁻⁴ C.

(iii) The charge stored in 3μF capacitor is 6 x 10⁻⁶ C.

<h3>Total capacitance of the circuit</h3>

The total capacitance of the circuit is determined by reolving the series capacitors separate and parallel capacitors separate as well.

<h3>C1 and C2 are in series </h3>

$\frac{1}{C_{12}} = \frac{1}{C_1 } + \frac{1}{C_2} \\\\\frac{1}{C_{12}} = \frac{1}{4 } + \frac{1}{4} \\\\\frac{1}{C_{12}} = \frac{1}{2} \\\\C_{12} = 2 \ \mu F$

<h3>C1 and C2 are parallel to C3</h3>

$C_{123} = C_{12} + C_3\\\\C_{123} = 2\ \mu F + 2\ \mu F \\\\C_{123} = 4 \ \mu F$

<h3>C(123) is series to C5 and C6</h3>

$\frac{1}{C_{t} } = \frac{1}{C_{123}} + \frac{1}{C_5} + \frac{1}{C_6} \\\\\frac{1}{C_{t} } = \frac{1}{4} + \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{t} } = \frac{12}{24} \\\\\frac{1}{C_{t} } = \frac{1}{2} \\\\C_t = 2 \ \mu F$

<h3>C7 and C8 are in series</h3>

$\frac{1}{C_{78}} = \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{78}} = \frac{2}{6} \\\\\frac{1}{C_{78}} =\frac{1}{3} \\\\C_{78} = 3 \ \mu F$

<h3>Total capaciatnce of the circuit</h3>

Ct + C(78) = 2 μF + 3 μF = 5 μF

<h3 /><h3>Total charge stored in the circuit</h3>

The total charge stored in the capacitor is calculated as follows;

Q = CV

Q = (5 x 10⁻⁶) x (20)

Q = 1 x 10⁻⁴ C

<h3>Charge stored in 3μF capacitor</h3>

Q = (3 x 10⁻⁶) x (20)

Q = 6 x 10⁻⁶ C

Learn more about capacitance of capacitor here: brainly.com/question/13578522

8 0

1 year ago

An 820 N Marine in basic training climbs a 12.0-m vertical rope at a constant speed in 8.00 s. What is her power output

sweet-ann [11.9K]

Answer:

1230 W

Explanation:

P = $\frac{W}{t}$ = $\frac{Force * distance}{time}$ = $\frac{820 N * 12.0 m}{8.00 s}$ = 1230 Watts

6 0

3 years ago

A student makes the following statement

Aliun [14]

Answer:

The current is not used up. The electrons flow through the entire circuit and this travel is the current. They flow until they are not charged anymore. That is also why the circuit must be closed or else electrons would escape not just light it up for a second then go out.

Explanation:

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3 years ago

If we decrease the distance an object moves we will (2 points)

Fantom [35]

Decrease the amount of force applied

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3 years ago

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