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3 years ago

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One 10.0 Ω resistor is wired in series with two 10.0 Ω resistors in parallel. A 45.0 V battery supplies power to the circuit. Wh

at is the total current through the circuit?

2 answers:

shutvik [7]3 years ago

7 0

You can chek this page "<span>Charges électriques - Courant électrique - Eduscol" the lesson is in frensh but you can traduct

</span>

RideAnS [48]3 years ago

4 0

The answer you are looking for 3.0 A

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How do you determine true displacement value of an object that has moved?

WITCHER [35]

Diceplacement is the distance an object has traveled in a certain direction
for example, if you were to walk North for 20m, then east for 40m, the <u>distance</u> you have traveled is 60m however your displacement is the distance between your starting position and your end position;
sqrt(20^2+40^2) = 44.7m
and because displacement is a vector, there needs to be a direction;
sin(theta)=40/44.7
theta=63.4 degrees East of North

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3 years ago

In an electric shaver, the blade moves back and forth over a distance of 2.0 mm in simple harmonic motion, with frequency 124 Hz

bagirrra123 [75]

Answer:

(a) A = 1 mm

(b) $V_{max}=0.77872 m/s$

(c) $a_{max}=606.4 m/s^{2}/tex]Explanation:Distance moved back and forth = 2 mm Frequency, f = 124 HzSo, amplitude is the half of the distance traveled back and forth. (a) So, amplitude, A = 1 mm(b) Angular frequency, ω = 2 π f = 2 x 3.14 x 124 = 778.72 rad/s The formula for the maximum speed is given by [tex]V_{max}=\omega \times A$

$V_{max}=778.72 \times 0.001$

$V_{max}=0.77872 m/s$

(c) The formula for the maximum acceleration is given by

$a_{max}=\omega ^{2}A$

$a_{max}=778.72 ^{2}\times 0.001$

[tex]a_{max}=606.4 m/s^{2}/tex]

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3 years ago

How much current must be applied across a 60 Ω light bulb filament in order for it to consume 55 W of power? Unserious answers w

sergij07 [2.7K]

Answer: current I = 0.96 Ampere

Explanation:

Given that the

Resistance R = 60 Ω

Power = 55 W

Power is the product of current and voltage. That is

P = IV ...... (1)

But voltage V = IR. From ohms law.

Substitutes V in equation (1) power is now

P = I^2R

Substitute the above parameters into the formula to get current I

55 = 60 × I^2

Make I^2 the subject of formula

I^2 = 55/60

I^2 = 0.92

I = sqr(0.92)

I = 0.957 A

Therefore, 0.96 A current must be applied.

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Hello!

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3 years ago