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3 years ago

Explain why light is referred to as electromagnetic radiation.

1 answer:

4 0

Answer: Light is a form of electromagnetic radiation because it has the ability to travel from one type of medium to another and can also carry some energy along with it. A light wave has two components, namely the electric field component as well as the magnetic field component that are both perpendicular to the propagating direction.

The visible light lies between the ultraviolet and the infrared lights, with wavelength from 400 nm to 700 nm respectively. This is collectively known as the visible spectrum of light.

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One advantage to using this form of circuit is that

777dan777 [17]

It has to be the last one because whenever lights are turned on it decreases because all lights are on at the same time. It's good to just have one light on. It doesn't use as much electricity.

8 0

2 years ago

A point charge q1q1 is held stationary at the origin. A second charge q2q2 is placed at point aa, and the electric potential ene

Yuri [45]

Explanation:

The given data is as follows.

$U_{a} = 5.4 \times 10^{-8} J$

$W_{/text{a to b}} = -1.9 \times 10^{-8} J$

Electric potential energy ( $U_{b}$ ) = ?

Formula to calculate electric potential energy is as follows.

$U_{b} = U_{a} - W_{/text{a to b}}$

= $5.4 \times 10^{-8} J - (-1.9 \times 10^{-8} J)$

= $7.3 \times 10^{-8} J$

Thus, we can conclude that the electric potential energy of the pair of charges when the second charge is at point b is $7.3 \times 10^{-8} J$ .

6 0

3 years ago

Un the way to the moon, the Apollo astro-

kherson [118]

Answer:

Distance = 345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

Second part

The distance between the Earth and this point is calculated as follows:

re = 3.84 108 - 38280860.6 = 345719139.4[m]

Now the acceleration can be found as follows:

$a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2} } \\a=3.33*10^{19} [m/s^2]$

6 0

3 years ago

What is the function of the lens of an eye

KATRIN_1 [288]

The lens changes the focal distance of the eye so that you can focus on images/objects.

8 0

3 years ago

Read 2 more answers

A 100-n force has a horizontal component of 80 n and a vertical component of 60 n. the force is applied to a box which rests on

iragen [17]

160 Joules

For this problem, we can ignore the vertical component of the applied force and focus on only the horizontal component of 80 N and since work is defined as force over distance, let's multiply the force by the distance: 80 N * 2.0 m = 160 Nm = 160 kg*m^2/s^2 = 160 Joules.

So the cart has a final kinetic energy of 160 Joules.

6 0

2 years ago