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2 years ago

Do electromagnetic waves require matter to travel?

Physics

1 answer:

VMariaS [17]2 years ago

6 0

Answer:

No.

Explanation:

Electromagnetic waves do not require a medium of matter to move through, electromagnetic waves are used in things like your cell phone and telecommunications.

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Is it possible for an object to have a charge of 4.8x10-9 C? Why or why not?

Gelneren [198K]

I have absolutely no clue

3 0

3 years ago

Many minerals, such as silver, are good conductors of heat and electricity. How could technology use this property to help ident

Aleks [24]

Answer:

This property could be used to create technologically-advanced tools or machines that could easily locate the mineral deposits.

Explanation:

Mineral deposits are hard to find, unless you have the skill or the proper tools in locating them. This is the reason why many people are mining in order to explore the different areas where they could find these deposits.

If one would consider the property of minerals, such as being good conductors of heat and electricity, then they could create a tool or machine that would aid in their exploration. Inventors could probably come up with a sensitive detector which signals when it reaches an area of high heat and electric conductivity. Since most minerals such as gold, silver, copper, galena, bornite and the like have this property, then miners will have a lesser amount of time looking for them.

If this technology will be implemented, though, regulation policy must be strictly implemented because it might lead to over-mining thus leading to the depletion of mineral deposits.

3 0

3 years ago

If a proton and an electron are released when they are 7.00×10−10 m apart (typical atomic distances), find the initial accelerat

Ugo [173]

Answer:

The acceleration of the proton is 2.823 x 10¹⁷ m/s²

The acceleration of the electron is 5.175 x 10²⁰ m/s²

Explanation:

Given;

distance between the electron and proton, r = 7 x 10⁻¹⁰ m

mass of proton, $m_p$ = 1.67 x 10⁻²⁷ kg

mass of electron, $m_e$ = 9.11 x 10⁻³¹ kg

The attractive force between the two charges is given by Coulomb's law;

$F = \frac{k(q_p)(q_e)}{r^2}$

where;

k is Coulomb's constant = 9 x 10⁹ Nm²/c²

$F = \frac{k(q_p)(q_e)}{r^2} \\\\F = \frac{(9*10^9)(1.602*10^{-19})(1.602*10^{-19})}{(7*10^{-10})^2} \\\\F = 4.714 *10^{-10} \ N$

Acceleration of proton is given by;

F = ma

$F = m_pa_p\\\\a_p = \frac{F}{m_p}\\\\a_p = \frac{4.714*10^{-10}}{1.67*10^{-27}}\\\\a_p = 2.823 *10^{17} \ m/s^2$

Acceleration of the electron is given by;

$F = m_ea_e\\\\a_e = \frac{F}{m_e}\\\\a_e = \frac{4.714*10^{-10}}{9.11*10^{-31}}\\\\a_e = 5.175 *10^{20} \ m/s^2$

7 0

3 years ago

Rank the tensions in the ropes, t1, t2, and t3, from smallest to largest, when the boxes are in motion and there is no friction

gizmo_the_mogwai [7]

AS T1,T2,T3 are the tensions in the ropes,assuming that there are Three blocks of mass 3m, 2m, and m.T3 is the string between 3m and 2m,T2 is the string between 2m and m ,T1 is the string attached to m thus T1 pulls the whole set of blocks along, so it must be the largest. T2 pulls the last two masses, but T3 only pulls the last mass, so T3 < T2 < T1.

5 0

3 years ago

An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.7

Kazeer [188]

Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

$v = \frac{I}{nqA}$

Where:

I: is the current = 4.50 A

n: is the number of electrons

q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C

A: is the cross-sectional area = 2.20x10⁻⁶ m²

We need to find the number of electrons:

$n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3}$

Now, we can find the drift speed:

$v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s$

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!

4 0

3 years ago