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murzikaleks [220]
3 years ago
15

How do electromagnetic waves affect you?

Physics
2 answers:
makvit [3.9K]3 years ago
7 0

Answer:

The main effect of radiofrequency electromagnetic fields is heating of body tissues.

Explanation:

Image result for How do electromagnetic waves affect you?

At low frequencies, external electric and magnetic fields induce small circulating currents within the body. ... The main effect of radiofrequency electromagnetic fields is heating of body tissues. There is no doubt that short-term exposure to very high levels of electromagnetic fields can be harmful to health.

Ivanshal [37]3 years ago
4 0

Answer:

At low frequencies, external electric and magnetic fields induce small circulating currents within the body. ... The main effect of radiofrequency electromagnetic fields is heating of body tissues. There is no doubt that short-term exposure to very high levels of electromagnetic fields can be harmful to health.

Explanation:

Found answer on google

hope it helps

please mark as brainliest

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Which element's atoms have the greatest average number of neutrons ?
Mkey [24]
I am pretty sure this is uranium. it has 140 neutrons. 
6 0
4 years ago
Read 2 more answers
A 7 ft tall person is walking away from a 20 ft tall lamppost at a rate of 5 ft/sec. Assume the scenario can be modeled with rig
aleksandr82 [10.1K]

Answer:

The rate of change of the shadow length of a person is 2.692 ft/s

Solution:

As per the question:

Height of a person, H = 20 ft

Height of a person, h = 7 ft

Rate = 5 ft/s

Now,

From Fig.1:

b = person's distance from the lamp post

a = shadow length

Also, from the similarity of the triangles, we can write:

\frac{a + b}{20} = \frac{a}{7}

a = \frac{7}{13}b

Differentiating the above eqn w.r.t t:

\frac{da}{dt} = \frac{7}{13}.\frac{db}{dt}

Now, we know that:

Rate = \frac{db}{dt} = 5\ ft/s

Thus

\frac{da}{dt} = \frac{7}{13}.\times 5 = 2.692\ ft/s

5 0
3 years ago
A plane travels at a speed of 205mph in still air. Flying with a tailwind, the plane is clocked over a distance of 1000 miles. F
vaieri [72.5K]

While plane is moving under tailwind condition it took time "t"

so here we will have

t = \frac{d}{v_{net}}

here net speed of the plane will be given as

v_{net} = v + v_w

t = \frac{1000}{205 + v_w}

similarly when it moves under the condition of headwind its net speed is given as

v_{net} = v - v_w

now time taken to cover the distance is 2 hours more

t + 2 = \frac{1000}{205 - v_w}

now solving two equations

\frac{1000}{205 + v_w} + 2 = \frac{1000}{205 - v_w}

solving above for v_w we got

v_w = 40.4 mph

6 0
3 years ago
A body of volume 2000 cm3 has a mass 4 kg. Find the density of the body. Will the body sink or float in water, if the density of
Brums [2.3K]

Answer:

2 g/cm^3, it will sink

Explanation:

The density of an object is given by

d=\frac{m}{V}

where

m is the mass of the object

V is its volume

For the body in the problem, we have

m = 4 kg = 4000 g

V=2000 cm^3

Therefore, its density is

d=\frac{4000}{2000}=2 g/cm^3

And the object will sink in water, because its density is larger than that of water, which is 1 g/cm^3. (an object sinks when its density is larger than that of water, otherwise it floats).

6 0
4 years ago
A projectile is launched horizontally from the top a 35.2m high cliff and lands a distance of 107.6m from the base of the cliff.
tankabanditka [31]

Answer:

v_o=40.14\ m/s

Explanation:

<u>Horizontal Launch </u>

It happens when an object is launched with an angle of zero respect to the horizontal reference. It's characteristics are:

  • The horizontal speed is constant and equal to the initial speed v_o
  • The vertical speed is zero at launch time, but increases as the object starts to fall
  • The height of the object gradually decreases until it hits the ground
  • The horizontal distance where the object lands is called the range

We have the following formulas

\displaystyle v_x=v_o

\displaystyle x=v_o.t

\displaystyle v_y=g.t

\displaystyle y=\frac{gt^2}{2}

Where v_o is the initial horizontal speed, v_y is the vertical speed, t is the time, g is the acceleration of gravity, x is the horizontal distance, and y is the height.

If we know the initial height of the object, we can compute the time it takes to hit the ground by using

\displaystyle y=\frac{gt^2}{2}

Rearranging and solving for t

\displaystyle 2y=gt^2

\displaystyle t^2=\frac{2\ y}{g}

\displaystyle t=\sqrt{\frac{2\ y}{g}}

We then replace this value in

\displaystyle x=v_o.t

To get

\displaystyle v_o=\frac{x}{t}

\displaystyle v_o=\frac{x}{\sqrt{\frac{2y}{g}}}

\displaystyle v_o=\sqrt{\frac{g}{2y}}.x

The initial speed depends on the initial height y=32.5 m, the range x=107.6 m and g=9.8 m/s^2. Computing v_o

\displaystyle v_o=\sqrt{\frac{9.8}{2(35.2)}}\ 107.6

The launch velocity is  

\boxed{v_o=40.14\ m/s}

7 0
3 years ago
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