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marishachu
3 months ago
6

Two Resistances R1 = 3 Ω and R2 = 6 Ω are connected in series with an ideal battery supplying a voltage of ∆ = 9 Volts. Sketch t

his circuit diagram. Now, replace the two resistors with an equivalent resistance R connected to the same battery. Sketch this circuit. (a) What is current I in R? (b) What is the potential difference V across R? Using this information, answer the following questions about the original, two-resistor circuit. (c) What is the current I1 in R1? (d) What is the current I2 in R2? (e) What is the potential difference V1 across R1? (f) What is in the potential difference V2 across R2? (g) How are V1 and V2 related to the battery voltage? Comparing the two circuits: (h) How are I1 and I2 related to I? (i) How are V1 and V2 related to ∆?
Two Resistances R1 = 3 Ω and R2 = 6 Ω are connected in parallel with an ideal battery supplying a voltage of ∆ =
9 Volts. Now, replace the two resistors with an equivalent resistance R connected to the same battery. Sketch this circuit. (a) What is current I in R? (b) What is the potential difference V across R? Using this information, answer the following questions about the original, two-resistor circuit. (c) What is the current I1 in R1? (d) What is the current I2 in R2? (e) What is the potential difference V1 across R1? (f) What is in the potential difference V2 across R2? (g) How are V1 and V2 related to the battery voltage? Comparing the two circuits: (h) How are I1 and I2 related to I? (i) How are V1 and V2 related to ∆?

Physics
1 answer:
Ostrovityanka [3.2K]3 months ago
4 0
For the requested solution, please refer to the attachment.
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Answer:

17 m/s south

Explanation:

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u_2 Velocity of skateboard

In this scenario, linear momentum is conserved

(m_1+m_2)v+m_1u_1+m_2u_2=0\\\Rightarrow u_2=-\dfrac{(m_1+m_2)v+m_1u_1}{m_2}\\\Rightarrow u_2=-\dfrac{(10+2)2+10\times 1}{2}\\\Rightarrow u_2=-17\ m/s

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1 month ago
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2 months ago
A 5.00-A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5 * 1028 free ele
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Answer:

a)n= 3.125 x 10^{19 electrones.

b)J= 1.515 x 10^{6 A/m²

c)V_{d =1.114 x 10^{4m/s

d) ver explicación

Explanation:

La corriente 'I' = 5A =>5C/s

diámetro 'd'= 2.05 x 10^{-3 m

radio 'r' = d/2 => 1.025 x 10^{-3 m

número de electrones 'n'= 8.5 x 10^{28}

a) La cantidad de electrones que pasan por la bombilla cada segundo se determina mediante:

I= Q/t

Q= I x t => 5 x 1

Q= 5C

Como sabemos que: Q= ne

donde e es la carga del electrón, es decir, 1.6 x 10^{-19C

n= Q/e => 5/ 1.6 x 10^{-19

n= 3.125 x 10^{19 electrones.

b) La densidad de corriente 'J' en el cable se calcula como

J= I/A => I/πr²

J= 5 / (3.14 x (1.025x 10^{-3)²)

J= 1.515 x 10^{6 A/m²

c) La velocidad típica 'V_{d' de un electrón se expresa como:

V_{d = \frac{J}{n|q|}

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V_{d =1.114 x 10^{4m/s

d) De acuerdo con estas ecuaciones,

J= I/A

V_{d = \frac{J}{n|q|} =\frac{I}{nA|q|}

Si utilizaras un cable de doble diámetro, ¿cuáles de las respuestas anteriores cambiarían? ¿Aumentarían o disminuirían?

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V - wind speed;
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When 999mm is added to 100m ______ is the result​
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Answer:

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Explanation:

That's my belief.

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