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natima [27]
3 years ago
15

Which of the following is an electromagnetic wave?

Physics
2 answers:
morpeh [17]3 years ago
5 0
Your answer will be Radio Waves . 

That seems to be the only to make sense. Hope that helps u 
Blizzard [7]3 years ago
5 0

Examples of Electromagnetic Waves. RADIO waves, Light waves, thermal radiation, X ray, visible light, microwave, infrared, gamma rays etc. are the example of electromagnetic waves. These waves together form the electromagnetic spectrum. (https://physics.tutorvista.com/waves/electromagnetic-waves.html)

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Verdich [7]

Answer:

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Explanation:

4 0
3 years ago
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Which of the following would not be useful in trying to correct spherical aberration a. using a combination lens made up of lens
Viefleur [7K]
A. Using a combination lens made up of lenses, each of which has a different index of refraction. Is the correct answer.
8 0
3 years ago
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What is the work required for a penguin to push a box 2 meters with a force of 8 newtons?
choli [55]

Work done is given by product of force and displacement due to that force

So here we will have

Work = Force \times displacement

here we know that

Force = 8 N

displacement = 2 m

Now work done is given as

W = 8\times 2

W = 16 J

so it will do 16 J work to move the box

3 0
3 years ago
A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

4 0
3 years ago
If V has a magnitude of 14 units and the same direction qs a vector 3i+6j+2k find v​
melomori [17]

Answer:

v = 6i + 12j + 4k

Explanation:

Find the magnitude of the direction vector.

√(3² + 6² + 2²) = 7

Normalize the direction vector.

3/7 i + 6/7 j + 2/7 k

Multiply by the magnitude of v.

v = 14 (3/7 i + 6/7 j + 2/7 k)

v = 6i + 12j + 4k

7 0
3 years ago
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