Please help me out if you want brainliest !!!!! ASAP easy

fredd [130]

We have here what is known as parallel combination of resistors.

Using the relation:

$\frac{1}{ r_{eff} } = \frac{1}{ r_{1} } + \frac{1}{ r_{2} } + \frac{1}{ r_{3} }.. . + \frac{1}{ r_{n} } \\$
And then we can turn take the inverse to get the effective resistance.

Where r is the magnitude of the resistance offered by each resistor.

In this case we have,
(every term has an mho in the end)
$\frac{1}{10000} + \frac{1}{2000} + \frac{1}{1000} \\ \\ = \frac{1}{1000} ( \frac{1}{10} + \frac{1}{2} + \frac{1}{1} ) \\ \\ = \frac{1}{1000} ( \frac{31}{20}) \\ \\ = \frac{31}{20000}$

To ger effective resistance take the inverse:
we get,
$\frac{20000}{31} \: ohm \\ = 645 .16 \: ohm$

The potential difference is of 9V.

So the current flowing using ohm's law,

V = IR

will be, 0.0139 Amperes.

7 0

2 years ago

A typical helium-neon laser found in supermarket checkout scanners emits 633-nm-wavelength light in a 1.0-mm-diameter beam with

Fofino [41]

Answer:

Eo = 9.796 x 10^2 N/C

Bo = 3.266 x 10^-6 T

Explanation:

Given

Wavelength λ = 633 nm

Diameter of the beam D = 1.0 mm

Power P = 1.0 mW

Solution

Radius of the beam r = D/2 = 0.5 mm = 0.0005 m

Area of cross section

$A = \pi r^{2} \\A = 3.15 \times 0.0005^{2}\\A = 7.58 \times 10^{-7} m^{2}\\$

Intensity

$I = \frac{P}{A} \\I = \frac{0.001}{7.85\times 10^{-7}} \\I = 1273.885 {W}/{m^{2} }$

Amplitude of Electric Field

$E_{o} = \sqrt{\frac{2I}{ \epsilon_{o}c } } \\E_{o} = \sqrt{\frac{2 \times 1273.88}{ 8.85 \times 10^{-12} \times 3 \times 10^{8} } }\\E_{o} = 9.796 \times 10^{2}N/C$

Amplitude of Magnetic Field

$B_{o} = \sqrt{\frac{2 \mu_{o}I}{c } } \\B_{o} = \sqrt{\frac{2 \times 4 \times \pi \times 10^{-7} \times 1273.88}{ 3 \times 10^{8} } }\\B_{o} = 3.266 \times 10^{-6} T$

5 0

3 years ago

The engine starter and a headlight of a car are connected in parallel to the 12.0-V car battery. In this situation, the headligh

stepladder [879]

Answer:

The total power they will consume in series is approximately 2.257 W

Explanation:

The connection arrangement of the headlight and the engine starter = Parallel to the battery

The voltage of the battery, V = 12.0 V

The power at which the headlight operates in parallel, $P_{headlight}$ = 38 W

The power at which the kick starter operates in parallel, $P_{kick \ starter}$ = 2.40 kW

We have;

P = V²/R

Where;

R = The resistance

V = The voltage = 12 V (The voltage is the same in parallel circuit)

For the headlight, we have;

R₁ = V²/ $P_{headlight}$ = 12²/38 = 72/19

R₁ = 72/19 Ω

For the kick starter, we have;

R₂ = V²/ $P_{kick \ starter}$ = 12²/2.4 = 60

R₂ = 60 Ω

When the headlight and kick starter are rewired to be in series, we have;

Total resistance, R = R₁ + R₂

Therefore;

R = ((72/19) + 60) Ω = (1212/19) Ω

The current flowing, I = V/R

∴ I = 12 V/(1212/19) Ω = (19/101) A

We note that power, P = I²R

In the series connection, we have;

$P_{headlight}$ = I² × R₁

∴ $P_{headlight}$ = ((19/101) A)² × 72/19 Ω = 1368/10201 W ≈ 0.134 W

The power at which the headlight operates in series, $P_{headlight, S}$ ≈ 0.134 W

$P_{kick \ starter}$ = ((19/101) A)² × 60 Ω = 21660/10201 W ≈ 2.123 W

The power at which the kick starter operates in series, $P_{kick \ starter, S}$ ≈ 2.123 W

The total power they will consume, $P_{Total}$ = $P_{headlight, S}$ + $P_{kick \ starter, S}$

Therefore;

$P_{Total}$ ≈ 0.134 W + 2.123 W = 2.257 W

4 0

2 years ago

4. When setting goals, you should do everything EXCEPT which of the following?

miv72 [106K]

Ask all of your friends what your goals should be. These are your goals, and they are for you only. They shouldn’t be what other people think they should be, they should be what you want them to be.

5 0

2 years ago

Indicate whether the statement is true or false. A hypothesis is not scientific if there exists no test for proving it wrong.

elena-s [515]

A. TRUE
A hypothesis must be based on scientific principles and rules in order to be experimented

5 0

3 years ago