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1 year ago

What is current of 12 ohm? help me pleasee

Physics

1 answer:

worty [1.4K]1 year ago

3 0

1 Ampere

Explanation:

1/R = 1/8 + 1/10 + 1/12
1/R = (30 + 24 + 20) / 240
1/R = 74 / 240
R = 240 / 37
R = 120/37 Ohms
We know that,
V = IR
I = V/R
I = 12 / (120/37)
I = 12 × 37/120
I = 37/10
I = 3.7 A
Now,
The current in 12 ohm resistor →
= 1 A
∴ The current in 12 ohm resistor is 1 ampere

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11 months ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

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3 years ago

A pilot wishes to flt his planne to an airport north of his current location

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Answer:

Time = 2758.62 seconds

Explanation:

Given the following data;

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Time = 2758.62 seconds

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2 years ago

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Answer:

Explanation:

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= 280 m / min

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total time = 250 min

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2 years ago

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Alenkinab [10]

Answer:

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