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2 years ago

Higher frequency electromagnetic waves have ?

Physics

1 answer:

Sergeeva-Olga [200]2 years ago

3 0

High frequency for the range of radio frequency electronmagnetic waves between 3 and 30 megahertz (MHz)

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In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.

leonid [27]

Answer:

(a): $F_e = 8.202\times 10^{-8}\ \rm N.$

(b): $F_g = 3.6125\times 10^{-47}\ \rm N.$

(c): $\dfrac{F_e}{F_g}=2.27\times 10^{39}.$

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053 $\times 10^{-9}$ m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges $q_1$ and $q_2$ respectively is given by

$F_e = \dfrac{k|q_1||q_2|}{r^2}$

where,

$k$ = Coulomb's constant = $9\times 10^9\ \rm Nm^2/C^2.$
$r$ = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, $q_1 = +1.6\times 10^{-19}\ C.$

The charge on the electron, $q_2 = -1.6\times 10^{-19}\ C.$

These two are separated by the distance, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

$F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.$

Part (b):

The gravitational force of attraction between two objects of masses $m_1$ and $m_1$ respectively is given by

$F_g = \dfrac{Gm_1m_2}{r^2}.$

where,

$G$ = Universal Gravitational constant = $6.67\times 10^{-11}\ \rm Nm^2/kg^2.$
$r$ = distance of separation between the masses.

For the given system,

The mass of proton, $m_1 = 1.67\times 10^{-27}\ kg.$

The mass of the electron, $m_2 = 9.11\times 10^{-31}\ kg.$

Distance between the two, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

$F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.$

The ratio $\dfrac{F_e}{F_g}$ :

$\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.$

6 0

3 years ago

Consider the plot below describing motion along a straight line with an initial position of 10 m. −4 −3 −2 −1 0 1 2 3 4 5 6 1 2

Ugo [173]

Answer:

$+1 m/s^2$

Explanation:

Acceleration is given by

$a=\frac{\Delta v}{\Delta t}$

where

$\Delta v$ is the change in velocity

$\Delta t$ is the time interval in which the change in velocity occurs

To find the acceleration at 1 second, we can take the data at t = 1 s and t = 2. We find:

$\Delta t = -3 -(-4) = 1 s$

$\Delta v = 2 - (1) = +1 m/s$

So, the acceleration is

$a=\frac{+1}{1}=+1 m/s^2$

4 0

3 years ago

Technology can best be defined as what?

jasenka [17]

B

Djdjsjdnkajdnsbsjsixhsjbsjsisahjsbxjsjsndnxnsjdd

7 0

2 years ago

Read 2 more answers

You go rock climbing with a pack that weighs 70 N and you reach a height of 30 m. how much work did you do to lift your pack? If

Vikentia [17]

When you climb, earth exerts gravitational force on pack in downward direction(pointing towards the center of earth).
In order to climb, you need to work against work done by gravity on the pack.
Hence work done by you = work done by gravity on pack
= Force x displacement = 70 x 30 = 2100 J.

So you need to do 2100 joules of work to lift your pack.

Power is the rate of work done.
Therefore power = work done by you/time(in seconds)
= 2100/600 =3.5 watts

8 0

3 years ago

Okay! Last question was a warm-up question. And now to get your brain thinking some more, how about another one?:

den301095 [7]

The correct answer is a fishhook

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2 years ago