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4 years ago

15

1. Find an example of electrical energy being converted to light energy.

1 answer:

Svetradugi [14.3K]4 years ago

6 0

Solar energy is the example of electrical energy

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How much heat is lost by 2.0 grams of water if the temperature drops from 31 °C to 29 °C? The specific heat of water is 4.184 J/

Elanso [62]

Given :

Mass of water, m = 2 grams.

The temperature of water drops from 31 °C to 29 °C .

The specific heat of water is 4.184 J/(g • °C).

To Find :

Amount of heat lost in this process.

Solution :

We know, heat lost is given by :

$Heat\ lost,H = ms( T_f - T_i)\\\\H = 2\times 4.184 \times ( 31 - 29 )\ J\\\\H = 16.736\ J$

Therefore, amount of heat lost in this process is 16.736 J.

4 0

3 years ago

A 259-kV power transmission line carrying 429 A is hung from grounded metal towers by ceramic insulators, each having a 0.71×109

ella [17]

Answer:

The resistance is $7.47\times10^{6}\ \Omega$ .

Explanation:

Given that,

Power = 259 kV

Current = 429 A

Resistance $R=0.71\times10^{9}\ Omega$

We need to calculate the current in each insulator

Using formula of current

$I=\dfrac{P}{R}$

Put the value into the formula

$I=\dfrac{259\times10^{3}}{0.71\times10^{9}}$

$I=3.64\times10^{-4}\ A$

So all 95 insulators are in parallel

We need to calculate the resistance

Using formula of resistance

$\dfrac{1}{R}=\sum_{i=1}^{95}\dfrac{1}{R_{i}}$

Put the value into the formula

$\dfrac{1}{R}=\dfrac{95}{0.71\times10^{9}}$

$\dfrac{1}{R}=1.338\times10^{-7}$

$R=\dfrac{1}{1.338\times10^{-7}}$

$R=7473841.5=7.47\times10^{6}\ \Omega$

Hence, The resistance is $7.47\times10^{6}\ \Omega$ .

8 0

4 years ago

DUE TODAY BEST ANSWER GET BRAINLIE

SCORPION-xisa [38]

Kinetic energy decreases if either the mass or the velocity of the object decreases or if both decrease. and Kinetic energy increases if either the mass or the velocity of the object increases or if both increase.

Explanation:

6 0

3 years ago

A red laser from the physics lab is marked as producing 632.8-nm light. When light from this laser falls on two closely spaced s

BARSIC [14]

Answer:

The wavelength is $\lambda_R = 649 *10^{-9}\ m$

Explanation:

From the question we are told that

The wavelength of the red laser is $\lambda_r = 632.8 \ nm = 632.8 *10^{-9}\ m$

The spacing between the fringe is $y_r = 6.00\ mm = 6.00*10^{-3} \ m$

The spacing between the fringe for smaller laser point is $y_R = 6.19 \ mm = 6.19 *10^{-3} \ m$

Generally the spacing between the fringe is mathematically represented as

$y = \frac{D * \lambda }{d}$

Here $D$ is the distance to the screen

and d is the distance of the slit separation

Now for both laser red light light and small laser point D and d are same for this experiment

So

$\frac{y_r}{\lambda_r} = \frac{D}{d}$

=> $\frac{y_r}{\lambda_r} = \frac{y_R}{\lambda_R}$

Where $\lambda_R$ is the wavelength produced by the small laser pointer

So

$\frac{6.0 *10^{-3}}{ 632.8*10^{-9}} = \frac{ 6.15 *10^{-9}}{\lambda_R}$

=> $\lambda_R = 649 *10^{-9}\ m$

7 0

3 years ago

We can reasonably model a 90-W incandescent lightbulb as a sphere 7.0cm in diameter. Typically, only about 5% of the energy goes

Ronch [10]

Answer:

292.3254055 W/m²

469.26267 V/m

$1.56421\times 10^{-6}\ T$

Explanation:

P = Power of bulb = 90 W

d = Diameter of bulb = 7 cm

r = Radius = $\frac{d}{2}=\frac{7}{2}=3.5\ cm$

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

c = Speed of light = $3\times 10^8\ m/s$

The intensity is given by

$I=\frac{P}{A}\\\Rightarrow I=\frac{90}{4\pi 0.035^2}\\\Rightarrow I=5846.50811\ W/m^2$

5% of this energy goes to the visible light so the intensity is

$I=0.05\times 5846.50811\\\Rightarrow I=292.3254055\ W/m^2$

The visible light intensity at the surface of the bulb is 292.3254055 W/m²

Energy density of the wave is

$u=\frac{1}{2}\epsilon_0E^2$

Energy density is also given by

$\frac{I}{c}=\frac{1}{2}\epsilon_0E^2\\\Rightarrow E=\sqrt{\frac{2I}{c\epsilon_0}}\\\Rightarrow E=\sqrt{\frac{2\times 292.3254055}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E=469.26267\ V/m$

The amplitude of the electric field at this surface is 469.26267 V/m

Amplitude of a magnetic field is given by

$B=\frac{E}{c}\\\Rightarrow B=\frac{469.26267}{3\times 10^8}\\\Rightarrow B=1.56421\times 10^{-6}\ T$

The amplitude of the magnetic field at this surface is $1.56421\times 10^{-6}\ T$

7 0

3 years ago