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3 years ago

What type of electromagnetic wave has a frequency of around 1012 Hz?

2 answers:

amm18123 years ago

4 0

Microwave frequencies range from about 109 Hz to the highest practical LC resonance at nearly 1012 Hz. Since they have high frequencies, their wavelengths are short compared with those of other radio waves—hence the name “microwave.” Microwaves can also be produced by atoms and molecules.

Pls Mark Brainliest if correct.

devlian [24]3 years ago

3 0

Answer: Hello, here is somthing that might help. Microwave frequencies range from about 109 Hz to the highest practical LC resonance at nearly 1012 Hz. Since they have high frequencies, their wavelengths are short compared with those of other radio waves—hence the name “microwave.” Microwaves can also be produced by atoms and molecules.

Explanation: Tell me if this helps o not :)

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What is the value of work done on an object when a 70 newton force moves it 9.0 meters in the same direction as the force in jou

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Work done= 70 *9 = 63J

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4 years ago

An unknown object is placed inside of a spherical container and dropped from an airplane. When empty, the spherical container ha

hichkok12 [17]

Answer:

The mass of unknown object is 8.62Kg

Explanation:

To develop this problem it is necessary to apply the equations related to the Drag force and the Force of Gravity.

For the given point, that is, the moment at which the terminal velocity is reached, the two forces equalize, that is,

$F_D =F_g$

By definition we know that the Drag force is defined as

$F_D= \frac{1}{2} C_d \rho A V^2$

Where,

$C_d =$ Drag coefficient

$\rho =$ Density

A =Cross-sectional Area

V = Velocity

In the other hand we have,

$F_g = (m_1 +m_2) g$

Where,

$m_1 =$ Mass of sphere

$m_2 =$ Mass of unknown object

Equating the two equations we have to

$(m_1 +m_2) g=\frac{1}{2} C_d \rho A V^2$

Re-arrange for m_2,

$m_2 = \frac{1}{2g} C_d \rho A V^2 -m_1$

Our values are given by,

$C_d = 0.5$

$\rho = 1.22Kg/m^3$

$V = 66.7m/s$

$m_1 = 3Kg$

$d= 32.7*10^{-2}m$

$r = 16.35*10^{-2}m$

Replacing in the equation we have,

$m_2 = \frac{1}{2(9.8)}(0.5) (1.22) (\pi*(16.35*10^{-2})^2)*66.7^2 -3$

$m_2 = 8.62Kg$

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4 0

3 years ago

Scientific work is currently underway to determine whether weak oscillating magnetic fields can affect human health. For example

nlexa [21]

Answer:

The maximum emf that can be generated around the perimeter of a cell in this field is $1.5732*10^{-11}V$

Explanation:

To solve this problem it is necessary to apply the concepts on maximum electromotive force.

For definition we know that

$\epsilon_{max} = NBA\omega$

Where,

N= Number of turns of the coil

B = Magnetic field

$\omega =$ Angular velocity

A = Cross-sectional Area

Angular velocity according kinematics equations is:

$\omega = 2\pi f$

$\omega = 2\pi*61.5$

$\omega =123\pi rad/s$

Replacing at the equation our values given we have that

$\epsilon_{max} = NBA\omega$

$\epsilon_{max} = NB(\pi (\frac{d}{2})^2)\omega$

$\epsilon_{max} = (1)(1*10^{-3})(\pi (\frac{7.2*10^{-6}}{2})^2)(123\pi)$

$\epsilon_{max} = 1.5732*10^{-11}V$

Therefore the maximum emf that can be generated around the perimeter of a cell in this field is $1.5732*10^{-11}V$

6 0

4 years ago

A certain atom has atomic number Z = 25 and atomic mass number A = 52. a. What is the approximate radius of the nucleus of this

wlad13 [49]

Answer:

a)The approximate radius of the nucleus of this atom is 4.656 fermi.

b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

Explanation:

$r=r_o\times A^{\frac{1}{3}}$

$r_o=1.25 \times 10^{-15} m$ = Constant for all nuclei

r = Radius of the nucleus

A = Number of nucleons

a) Given atomic number of an element = 25

Atomic mass or nucleon number = 52

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$r=4.6656\times 10^{-15} m=4.6656 fm$

The approximate radius of the nucleus of this atom is 4.656 fermi.

b) $F=k\times \frac{q_1q_2}{a^2}$

k= $9\times 10^9 N m^2/C^2$ = Coulombs constant

$q_1,q_2$ = charges kept at distance 'a' from each other

F = electrostatic force between charges

$q_1=+1.602\times 10^{-19} C$

$q_2=+1.602\times 10^{-19} C$

Force of repulsion between two protons on opposite sides of the diameter

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$F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}$

$F=2.6527 N$

The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

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3 years ago

Explain how Law 1 applies to the image to the left.

alisha [4.7K]

Answer:

Explanation:

3 0

3 years ago

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What type of electromagnetic wave has a frequency of around 1012 Hz?​

What type of electromagnetic wave has a frequency of around 1012 Hz?