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3 years ago

What are the dangers of microwaves in the Em spectrum?

Physics

2 answers:

Ray Of Light [21]3 years ago

8 0

The answer to this one is Electromagnetic radiation because it is dangerous.

SIZIF [17.4K]3 years ago

6 0

Prolonged exposure to significant levels of microwaves is known to cause "cataracts" in your eyes, which is a clouding of the lens preventing you from seeing clearly. People who work on aircraft carrier decks wear special suits which reflect microwaves, to avoid being "cooked" by the powerful radar units in modern military planes. But that's to guard against powerful radar - not mobile phones or wifi. Some research in the past has indicated that the small doses of microwaves from mobile phones might affect parts of your brain - after all, you're holding the transmitter right by your head. Other research is inconclusive, although there is a feeling that you're more vulnerable if you're young and your brain is still growing.

Hope this answer helps! feel free to ask any additional questions :)

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3 years ago

A long, thin rod parallel to the y-axis is located at x = - 1 cm and carries a uniform positive charge density λ = 1 nC/m . A se

zheka24 [161]

Answer:

The electric field at origin is 3600 N/C

Solution:

As per the question:

Charge density of rod 1, $\lambda = 1\ nC = 1\times 10^{- 9}\ C$

Charge density of rod 2, $\lambda = - 1\ nC = - 1\times 10^{- 9}\ C$

Now,

To calculate the electric field at origin:

We know that the electric field due to a long rod is given by:

$\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}$

Also,

$\vec{E} = \frac{2K\lambda }{R}$ (1)

where

K = electrostatic constant = $\frac{1}{4\pi \epsilon_{o} R}$

R = Distance

$\lambda$ = linear charge density

Now,

In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.

At x = - 1 cm = - 0.01 m:

Using eqn (1):

$\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C$

$\vec{E} = 1800\ N/C$ (towards)

Now, at x = 1 cm = 0.01 m :

Using eqn (1):

$\vec{E'} = \frac{2\times 9\times 10^{9}\times - 1\times 10^{- 9}}{0.01} = - 1800\ N/C$

$\vec{E'} = 1800\ N/C$ (towards)

Now, the total field at the origin is the sum of both the fields:

$\vec{E_{net}} = 1800 + 1800 = 3600\ N/C$

7 0

3 years ago

A 500-Ω resistor, an uncharged 1.50-μF capacitor, and a 6.16-V emf are connected in series. (a) What is the initial current? (b)

Alexeev081 [22]

Answer:

a) 0.01232 A

b) 0.00075 s = 0.75 ms

c) 0.0045323 A = 4.532 mA

d) 3.894 V

Explanation:

R = 500 Ω

V = 6.16 V

C = 1.50 μF

Let Vs be the voltage of the emf source

Let Vc be the voltage across the capacitor at any time

a) Current flows as a result of potential difference between two points. So, the current flows according to difference in voltage between the emf source and the capacitor.

At time t = 0,

There is no voltage on the capacitor; Vc = 0 V

Current in the circuit is given by

I = (Vs - Vc)/R

I = (6.16 - 0)/500

I = 0.01232 A

b) Time constant for an RC circuit is given by τ

τ = RC = (500) (1.5 × 10⁻⁶) = 0.00075 s

c) The current decay in an RC circuit (called decay because the current in the circuit starts to fall as the capacitor's voltage rises as the capacitor charges) is given by

I = I₀ e⁻ᵏᵗ

where k = (1/τ)

I₀ = Current in the circuit at t = 0 s; I₀ = 0.01232 A

At t = τ = 0.00075 s, kt = (τ/τ) = 1

I = 0.01232 e⁻¹ = 0.0045323 A = 4.532 mA

d) The voltage for a charging capacitor is given by

Vc = Vs (1 - e⁻ᵏᵗ)

where k = (1/τ)

At t = τ = 0.00075 s, Vc = ?, Vs = 6.16 V, kt = 1

Vc = 6.16 (1 - e⁻¹) = 6.16 (0.6321)

Vc = 3.894 V

4 0

3 years ago

Read 2 more answers