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3 years ago

Which of the following best describes why we say that light is an electromagnetic wave?

Physics

1 answer:

Kaylis [27]3 years ago

8 0

Answer: The passage of a light wave can cause electrically charged particles to move up and down.

Explanation:

Electromagnetic waves are transversal waves, they are a combination of oscillating electric and magnetic fields, which propagate through space carrying energy from one place to another.

This means the oscillation of the wave occurs in the transversal direction to its propagation. In addition, electromagnetic waves are spread thanks to the electromagnetic fields produced by moving electric charges.

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A car of mass 800kg travels a distance of 40m at constant speed in a duration of 2.0s. The car exerts a forward force of 15kN.

Alex17521 [72]

W = F × s

W = 15kN × 40 m

W = 15.000 N × 40 m

W = 600.000 J

P = W/t

P = 600.000 J/2 s

P = 300.000 Watt

P = 300kWatt

#LearnWithEXO

6 0

3 years ago

Student swings a small rubber stopper attached to a string over her head in a horizontal, circular path. The string is 1.50 mete

harkovskaia [24]

Answer:

v = 18.84 m/s

Explanation:

Given that,

The length of the string, r = 1.5 m (it will act as radius)

The rubber stopper makes 120 complete circles every minute.

Since, 1 minute = 60 seconds

It means, its frequency is 2 circles every second.

Let we need to find the average speed of the rubber stopper. It can be calculated as follows :

$v=\dfrac{d}{T}$

d is distance, $d=2\pi r$ and 1/T = f (frequency)

$v=2\pi rf\\\\=2\pi \times 1.5\times 2\\\\=18.84\ m/s$

So, the average speed of the rubber stopper is 18.84 m/s.

4 0

3 years ago

A concave mirror has a focal length of 13.5 cm. This mirror forms an image located 37.5 cm in front of the mirror. Find the magn

77julia77 [94]

Explanation:

It is given that,

Focal length of the concave mirror, f = -13.5 cm

Image distance, v = -37.5 cm (in front of mirror)

Let u is the object distance. It can be calculated using the mirror's formula as :

$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$

$\dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{v}$

$\dfrac{1}{u}=\dfrac{1}{(-13.5)}-\dfrac{1}{(-37.5)}$

u = -21.09 cm

The magnification of the mirror is given by :

$m=\dfrac{-v}{u}$

$m=\dfrac{-(-37.5)}{(-21.09)}$

m = -1.77

So, the magnification produced by the mirror is (-1.77). Hence, this is the required solution.

7 0

3 years ago

A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur

muminat

Answer:

a) E = 0

b) $E = \dfrac{k_e \cdot q}{ r^2 }$

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

From which we have;

$E \cdot A = \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}} = \dfrac{0}{\varepsilon _{0}} = 0$

E = 0/A = 0

E = 0

b) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

$E \cdot A = \dfrac{+q }{\varepsilon _{0}}$

$E = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}$

By Gauss theorem, we have;

$E\oint dS = \dfrac{q}{\varepsilon _{0}}$

Therefore, we get;

$E \cdot (4 \cdot \pi \cdot r^2) = \dfrac{q}{\varepsilon _{0}}$

The electrical field outside the spherical shell

$E = \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }= \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }$

$k_e= \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }$

Therefore, we have;

$E = \dfrac{k_e \cdot q}{ r^2 }$

5 0

3 years ago

An equilibrium constant is not changed by a change in pressure <br> a. True<br> b. False

Umnica [9.8K]

Hi There! :)

An equilibrium constant is not changed by a change in pressurea. True
b. False

False! :P

7 0

3 years ago

Read 2 more answers