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stealth61 [152]

3 years ago

10

If 100 J of electrical energy enter the bulb and 5 J of light energy leave the bulb, how many joules of heat energy leave the bu

lb?

1 answer:

Novay_Z [31]3 years ago

6 0

As per energy conservation we know that

Energy enter into the bulb = Light energy + Thermal energy

so now we have

energy enter into the bulb = 100 J

Light energy = 5 J

now from above equation we have

$100 = 5 + heat$

$Heat = (100 - 5) J$

$Heat = 95 J$

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What is the reflectivity of a glass surface (n =1.5) in air (n = 1) at an 45° for (a) S-polarized light and (b) P-polarized ligh

Goshia [24]

Answer:

a) $R_s = 0.092$

b) $R_p = 0.085$

Explanation:

given,

n =1.5 for glass surface

n = 1 for air

incidence angle = 45°

using Fresnel equation of reflectivity of S and P polarized light

$R_s=\left | \dfrac{n_1cos\theta_i-n_2cos\theta_t}{n_1cos\theta_i+n_2cos\theta_t} \right |^2\\R_p=\left | \dfrac{n_1cos\theta_t-n_2cos\theta_i}{n_1cos\theta_t+n_2cos\theta_i} \right |^2$

using snell's law to calculate θ t

$sin \theta_t = \dfrac{n_1sin\theta_i}{n_2}=\dfrac{sin45^0}{1.5}=\dfrac{\sqrt{2}}{3}$

$cos \theta_t =\sqrt{1-sin^2\theta_t} = \dfrac{sqrt{7}}{3}$

a) $R_s=\left | \dfrac{\dfrac{1}{\sqrt{2}}-\dfrac{1.5\sqrt{7}}{3}}{\dfrac{1}{\sqrt{2}}+\dfrac{1.5\sqrt{7}}{3}} \right |^2$

$R_s = 0.092$

b) $R_p=\left | \dfrac{\dfrac{\sqrt{7}}{3}-\dfrac{1.5}{\sqrt{2}}}{\dfrac{\sqrt{7}}{3}+\dfrac{1.5}{\sqrt{2}}} \right |^2$

$R_p = 0.085$

3 0

3 years ago

One of the foci for the moon's orbit would be the

suter [353]

The question is oversimplified, and pretty sloppy.

Relative to the Earth . . .
The Moon is in an elliptical orbit around us, with a period of
27.32... days, and with the Earth at one focus of the ellipse.

Relative to the Sun . . .
The Moon is in an elliptical orbit around the Sun, with a period
of 365.24... days, and with the Sun at one focus of the ellipse,
and the Moon itself makes little dimples or squiggles in its orbit
on account of the gravitational influence of the nearby Earth.

I'm sorry if that seems complicated. You know that motion is
always relative to something, and the solar system is not simple.

5 0

3 years ago

Read 2 more answers

I need a short answer ?

mars1129 [50]

Answer:

Explanation:

7a) t = d/v = 100/45cos14.5 = 2.29533...= 2.30 s

7b) h = ½(9.81)(2.29533/2)² = 6.46056... = 6.45 m

or

h = (45sin14.5)² / (2(9.81)) = 6.47 m

which rounds to the same 6.5 m when limiting to the two significant digits of the initial velocity.

7 0

2 years ago

How much can having a lighted candle increase the temperature inside the shelter?

oksian1 [2.3K]

A lighted candle produces heat however not as much heat as a heater or the sun would.

8 0

3 years ago

A 2 kg ball is moving 3 m/s when it starts rolling up a hill.

AURORKA [14]

Answer:

the height reached is = 0.458 [m]

Explanation:

We need to make a sketch of the ball and see the location of the reference point where the potential energy is zero. But the kinetic energy will be defined by the following expression:

$Ek=\frac{1}{2} *m*v^{2} \\where:Ek= kinetic energy [J]\\m = mass of the ball [kg]\\v = velocity of the ball [m/s]$

Replacing the values on the equation we have:

$Ek=\frac{1}{2}*(2)*(3^{2} )\\ Ek=9[J]\\$

This kinetic energy will be transformed in potential energy in the moment when the ball starts to rolling up. Therefore the maximum height reached by the ball depends of the initial velocity given to the ball.

$Ek=Ep\\where\\Ep=potential energy [J]\\Ep=m*g*h\\where\\g=gravity = 9.81[m/s^2]\\h=height reached [m]\\$

Now we have:

$h=\frac{Ep}{m*g} \\h=\frac{9}{2*9.81} \\\\h=0.45 [m]$

In that moment when the ball reach the 0.45 [m] the potencial energy will be maximum and equal to the kinetic energy when the ball has a velocity of 3[m/s]

6 0

3 years ago