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sattari [20]
3 years ago
6

What are examples of devices that use electromagnetic waves? Check all that apply.

Physics
2 answers:
yKpoI14uk [10]3 years ago
5 0

Well, there aren't actually ANY that apply, because you haven't listed any.  But I've lived a long time, and I remember hearing about electromagnetic waves and things that use them, so I can list a FEW of them for you:

-- radios

-- TVs

-- garage-door openers

-- TV remotes

-- cell phones

-- smart phones

-- GPS

-- walkie-talkies

-- car headlights

-- lava lamps

-- toasters

-- LEDs

-- light bulbs

-- fluorescent light tubes

-- police radios

-- Doppler weather radars

-- CB radios

-- ham radios

-- neon signs

-- eyeglasses

-- microscopes

-- telescopes

-- gas stoves

-- electric stoves

-- wood stoves

-- microwave ovens

-- tanning beds

-- cameras

-- lasers

-- CD recorders and players

-- DVD recorders and players

-- Bluray recorders and players

-- movie cameras

-- movie projectors

-- reading lamps

-- candles

-- whale-oil lamps

-- kerosene lanterns

-- flashlights

-- campfires

-- coffee percolators

-- heat lamps

-- cordless phones

These are just the ones I can think of right now off the top of my head.  There are a lot of others.

ratt3 years ago
0 0

its A, B, C, and E on edg:
FM radios
microwaves
TV remote controls
X-rays

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dolphi86 [110]

Answer:

d: 44m

Δd:34  

Explanation:

d: 23+5+16= 44m

Δd: 23-5+16= 34m

4 0
3 years ago
Each plate of an air-filled parallel-plate capacitor has an area of 45.0 cm2, and the separation of the plates is 0.080 mm. A ba
maw [93]

Answer:

Option (e)

Explanation:

A = 45 cm^2 = 0.0045 m^2, d = 0.080 mm = 0.080 x 10^-3 m,

Energy density = 100 J/m

Let Q be the charge on the plates.

Energy density = 1/2 x ε0 x E^2

100 = 0.5 x 8.854 x 10^-12 x E^2

E = 4.75 x 10^6 V/m

V = E x d

V = 4.75 x 10^6 x 0.080 x 10^-3 = 380.22 V

C = ε0 A / d

C = 8.854 x 10^-12 x 45 x 10^-4 / (0.080 x 10^-3) = 4.98 x 10^-10 F

Q = C x V = 4.98 x 10^-10 x 380.22 = 1.9 x 10^-7 C

Q =  190 nC

3 0
3 years ago
Wich advantage of reproduction does the graph shown
inn [45]

Answer:

No photo or graph is there

Explanation:

4 0
2 years ago
A parallel-plate capacitor is formed from two 1.6 cm -diameter electrodes spaced 2.8 mm apart. The electric field strength insid
lidiya [134]

Answer:

12 nC

Explanation:

Capacity of the parallel plate capacitor

C = ε₀ A/d

ε₀ is constant having value of 8.85 x 10⁻¹² , A area of plate , d is distance between plate

Area of plate = π r²

= 3.14 x (0.8x 10⁻²)²

= 2 x 10⁻⁴

C = ( 8.85 X 10⁻¹² x 2 x 10⁻⁴ ) / 2.8 x 10⁻³

= 7.08 x 10⁻¹³

Potential difference between plate = field strength x distance between plate

= 6 x 10⁶ x 2.8 x 10⁻³

= 16.8 x 10³ V

Charge on plate = CV

=7.08 x 10⁻¹³ X 16.8 X 10³

11.9 X 10⁻⁹ C

12 nC .

8 0
3 years ago
Please help with this problem. Thank you.​
4vir4ik [10]

Answer:

Explanation:

F = mω²R

F = 15(2π/8.5)²(7.8)

F = 63.93044788...

F = 63.9 N

answer a) is the closest. No idea how they got a value that low unless they used a poor approximation for π.

4 0
3 years ago
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