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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. <span>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 :)</span>

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My Notes Ask Your Teacher 6. Six blocks with different masses, m, each start from rest at the top of smooth, frictionless inclin

Tresset [83]

Answer:

Kf > Ka = Kb > Kc > Kd > Ke

Explanation:

We can apply

E₀ = E₁

where

E₀: Mechanical energy at the beginning of the motion (top of the incline)

E₁: Mechanical energy at the end (bottom of the incline)

then

K₀ + U₀ = K₁ + U₁

If v₀ = 0 ⇒ K₀

and h₁ = 0 ⇒ U₁ = 0

we get

U₀ = K₁

U₀ = m*g*h₀ = K₁

we apply the same equation in each case

a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J

d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J

e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J

f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J

finally, we can say that

Kf > Ka = Kb > Kc > Kd > Ke

8 0

3 years ago

The specialty of an athlete on the women's track team is the pole vault. She has a mass of 48 kg and her approach speed is 8.9 m

Lyrx [107]

Answer:

H = 3.9 m

Explanation:

mass (m) = 48 kg

initial velocity (initial speed) (U) = 8.9 m/s

final velocity (V) = 1.6 m/s

acceleration due to gravity (g) = 9.8 m/s^{2}

find the height she raised her self to as she crosses the bar (H)

from energy conservation, the change in kinetic energy = change in potential energy

0.5m(V^{2} - [test]U^{2}[/tex]) = mg(H-h)

where h = initial height = 0 since she was on the ground

the equation becomes

0.5m(V^{2} - [test]U^{2}[/tex]) = mgH

0.5 x 48 x (1.6^{2} - [test]8.9^{2}[/tex]) = 48 x 9.8 x H

-1839.6 = 470.4 H (the negative sign indicates a decrease in kinetic energy so we would not be making use of it further)

H = 3.9 m

4 0

4 years ago

A student is trying to determine the acceleration of a feather as she drops it to the ground. if the student is looking to achie

Anna [14]

The coordinate system should have the origin at the point where the feather is dropped and the downward direction is to be taken as positive.

All falling bodies experience acceleration towards the center of the Earth due to the force of gravitational attraction exerted on the object by the Earth. A feather, when dropped experiences an acceleration in the downward direction. Since the acceleration of the feather is in the downward direction, a feather, when dropped with zero initial velocity, has its velocity vector directed in the direction of its acceleration.

If the downward direction is taken as positive, the falling feather can be said to have a positive velocity and a positive acceleration.

5 0

3 years ago

car 2 has a mass of 150 kg and moves westward towards car 3 at a velocity of 2.2 m/s. car 3 has a mass of 265 kg and moves eastw

sergejj [24]

Answer:

The force of car 3 on car 2 ≈ 1810.82 N

Explanation:

The equation for the change in momentum of the two cars are;

Conservation of linear momentum

150( 2.2 - v) = 265(1.5-v)

150 × 2.2 - 265×1.5 = (150+265)v

150 × 2.2 - 265×1.5 = -67.5 = 415×v

∴ v = -67.5/415 = -0.1627 m/s West = 0.1627 m/s East

The impulse of the net force is the amount of momentum change experienced given by the equation;

Impulse force = $m \times v_f$ - $m \times v_0$

Where;

$v_f$ = The final velocity

$v_0$ = The initial velocity

For the the 265 kg mass, we have;

$v_f$ = 0.1627 m/s

$v_0$ = 1.5 m/s

Which gives the impulse a s F×Δt = 265×0.1627 - 265×1.5 = -354.38 kg·m/s

The change in kinetic energy of the collision = 1/2×265×(0.1627^2 - 1.5^2) =-294.62 J

Whereby the distance moved in one second is 0.1627 m, we have;

Work done = Force × Distance = Force × 0.1627 = 294.62

Force = 294.62/0.1627 = 1810.82 N.

8 0

3 years ago

Suppose Group A runs their experiment 3 times and calculates that the best estimate of the distance slid by red blocks is 15 + 3

Alex73 [517]

Sorry I just need points

7 0

3 years ago