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2 years ago

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(Brainliest) How are radio waves used in cell phone wireless communication technology?

1 answer:

poizon [28]2 years ago

5 0

Radio waves are used in cell phone wireless communication by this: Radio waves are released from the cell phone and travel to satellites in the stratosphere. After they reach the satellite, they are redirected to the recipient of the call/text.

Hope this helps! Can I have brainliest please?

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A 65.0-kg woman steps off a 10.0-m diving platform and drops straight down into the water. 1) If she reaches a depth of 3.20 m,

timurjin [86]

Answer:

F=2627.6N

Explanation:

The work done by this resistive force while traveling a distance <em>d</em> underwater would be:

$W=F.d=-Fd$

where the minus sign appears because the force is upwards and the displacement downwards.

This work is equal to the change of mechanical energy. At the diving plataform and underwater, when she stops moving, the woman has no kinetic energy, so all can be written in terms of her total change of gravitational potential energy:

$W=\Delta E=U_f-U_i=mgh_f-mgh_i=mg(h_f-h_i)$

Putting all together:

$F=-\frac{W}{d}=-\frac{mg(h_f-h_i)}{d}=-\frac{(65kg)(9.8m/s^2)(-3.2m-10m)}{3.2m}=2627.6N$

7 0

2 years ago

Which element could provide one atom to make an ionic bond with nitrogen?

aev [14]

Answer:

boron

aluminum

gallium

indium

thallium

Explanation:

Any of these could work. Nitrogen has 5 valence electrons so you just needed to pick an element that has 3 valence electrons that nitrogen could borrow. This periodic table shows valence electron counts:

8 0

2 years ago

A ball is shot at an angle of 45 degrees into the air with initial velocity of 46 ft/sec. Assuming no air resistance, how high d

mestny [16]

Answer:

5.02 m

Explanation:

Applying the formula of maximum height of a projectile,

H = U²sin²Ф/2g...................... Equation 1

Where H = maximum height, U = initial velocity, Ф = angle, g = acceleration due to gravity.

Given: U = 46 ft/sec = 14.021 m/s, Ф = 45°

Constant: g = 9.8 m/s²

Substitute these values into equation 1

H = (14.021)²sin²45/(2×9.8)

H = 196.5884×0.5/19.6

H = 5.02 m.

Hence the ball goes 5.02 m high

8 0

2 years ago

Two observers are 300 ft apart on opposite sides of a flagpole. The angles of elevation from the observers to the top of the pol

valentina_108 [34]

Answer:

h = 48.077 ft

Explanation:

given,

distance between two observer = 300 ft

angle of elevation to top pole = 16° and 20°

height of the flagpole = ?

now,

Let h be the height of the flagpole

Let x be the distance of the pole

$tan 16^0 = \dfrac{h}{x}$

$x =\dfrac{h}{tan 16^0}$

now,

again applying

$tan 20^0 = \dfrac{h}{300-x}$

$300-x=\dfrac{h}{tan 20^0}$

$300-3.49 h=2.75 h$

$6.24h = 300$

h = 48.077 ft

3 0

2 years ago

Objects 1 and 2 attract each other with a gravitational force

snow_tiger [21]

Answer:

F' = 162 units

Explanation:

The gravitational force of attraction between the two objects is given by Newton's Gravitational law through the following formula:

$F = \frac{Gm_{1}m_{2}}{r^{2}}\\\\$

where,

F = gravitational force = 18 units

G = Gravitational Constant

m₁ = mass of object 1

m₂ = mass of object 2

r = distance between objects

Therefore,

$18 = \frac{Gm_{1}m_{2}}{r^{2}}------ eqn (1)\\\\$

Now, if we change the value of distance to one-third of original value, then:

r' = r/3

$F' = \frac{Gm_{1}m_{2}}{(\frac{r}{3})^{2}}\\\\F' = (9)(\frac{Gm_{1}m_{2}}{r^{2}})$

using eqn (1):

F' = 9(18 units)

<u>F' = 162 units</u>

7 0

2 years ago