Examples of black body radiators? Please

In this circuit (see picture), which resistor will draw the least power?

Basile [38]

A few different ways to do this:

Way #1:
The current in the series loop is (12 V) / (total resistance) .
(Turns out to be 2 Amperes, but the question isn't asking for that.)

In a series loop, the current is the same at every point, so it's
the same current through each resistor.

The power dissipated by a resistor is (current)² · (resistance),
and the current is the same everywhere in the circuit, so the
smallest resistance will dissipate the least power. That's R1 .

And by the way, it's not "drawing" the most power. It's dissipating it.

Way #2:
Another expression for the power dissipated by a resistance is

(voltage across the resistance)² / (resistance) .

In a series loop, the voltage across each resistor is

[ (individual resistance) / (total resistance ] x battery voltage.

So the power dissipated by each resistor is

(individual resistance)² x [(battery voltage) / (total resistance)²]

This expression is smallest for the smallest individual resistance.
(The other two quantities are the same for each individual resistor.)
So again, the least power is dissipated by the smallest individual resistance.
That's R1 .

Way #3: (Einstein's way)
If we sat back and relaxed for a minute, stared at the ceiling, let our minds
wander, puffed gently on our pipe, and just daydreamed about this question
for a minute or two, we might have easily guessed at the answer.

===> When you wire up a battery and a light bulb in series, the part
that dissipates power, and gets so hot that it radiates heat and light, is
the light bulb (some resistance), not the wire (very small resistance).

3 0

2 years ago

The astronomical unit (AU) is defined as the mean center-to-center distance from Earth to the Sun, namely 1.496x10^(11) m. The p

Rudiy27

Answer:

a) How many parsecs are there in one astronomical unit?

$4.85x10^{-6}pc$

(b) How many meters are in a parsec?

$3.081x10^{16}m$

(c) How many meters in a light-year?

$9.46x10^{15}m$

(d) How many astronomical units in a light-year?

$63325AU$

(e) How many light-years in a parsec?

3.26ly

Explanation:

The parallax angle can be used to find out the distance using triangulation. Making a triangle between the nearby star, the Sun and the Earth, knowing that the distance between the Earth and the Sun ( $1.496x10^{11} m$ ) is defined as 1 astronomical unit:

$\tan{p} = \frac{1AU}{d}$

Where d is the distance to the star.

Since p is small it can be represent as:

$p(rad) = \frac{1AU}{d}$ (1)

Where p(rad) is the value of in radians

However, it is better to express small angles in arcseconds

$p('') = p(rad)\frac{180^\circ}{\pi rad}.\frac{60'}{1^\circ}.\frac{60''}{1'}$

$p('') = 2.06x10^5 p(rad)$

$p(rad) = \frac{p('')}{2.06x10^5}$ (2)

Then, equation 2 can be replace in equation 1:

$\frac{p('')}{2.06x10^5} = \frac{1AU}{d}$

$\frac{d}{1AU} = \frac{2.06x10^5}{p('')}$ (3)

From equation 3 it can be see that $1pc = 2.06x10^5 AU$

a) How many parsecs are there in one astronomical unit?

$1AU . \frac{1pc}{2.06x10^5AU}$ ⇒ $4.85x10^{-6}pc$

(b) How many meters are in a parsec?

$2.06x10^{5}AU . \frac{1.496x10^{11}m}{1AU}$ ⇒ $3.081x10^{16}m$

(c) How many meters in a light-year?

To determine the number of meters in a light-year it is necessary to use the next equation:

$x = c.t$

Where c is the speed of light ( $c = 3x10^{8}m/s$ ) and x is the distance that light travels in 1 year.

In 1 year they are 31536000 seconds

$x = (3x10^{8}m/s)(31536000s)$

$x = 9.46x10^{15}m$

(d) How many astronomical units in a light-year?

$9.46x10^{15}m . \frac{1AU}{1.496x10^{11}m}$ ⇒ $63325AU$

(e) How many light-years in a parsec?

$2.06x10^{5}AU . \frac{1ly}{63235AU}$ ⇒ $3.26ly$

5 0

3 years ago

Which of the following circuits can be used to measure the resistance of the heating element, shown as a resistor in the diagram

Wewaii [24]

In order to measure the resistance in the circuit, we need to know the voltage V and the current I in the circuit, this way we can calculate the resistance using the formula:

$R=\frac{V}{I}$

In order to calculate the current, we can use an amperemeter that must be in series with the circuit, this way it will not affect the circuit.

And in order to calculate the voltage, we can use a voltmeter that must be in parallel with the resistance, this way it will not affect the circuit.

The correct option that shows an amperemeter in series and a voltmeter in parallel is the fourth option.

8 0

1 year ago

A young's interference experiment is performed with blue-green laser light. the separation between the slits is 0.500 mm, and th

fgiga [73]

What class is this for?

3 0

3 years ago

Two boxes are 8 cm apart. Which of the following should Janet do to decrease the gravitational force between the boxes?

OlgaM077 [116]

Answer:

the answer is 2.

Explanation:

4 0

2 years ago