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Svetradugi [14.3K]
3 years ago
13

If two such generic humans each carried 1.0 coulomb of excess charge, one positive and one negative, how far apart would they ha

ve to be for the electric attraction between them to equal their 700 n weight?
Physics
1 answer:
Anuta_ua [19.1K]3 years ago
3 0
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Physical science help!
Scrat [10]

Answer:

Let's begin by explaining that the <u>electromagnetic spectrum</u> is the set of electromagnetic radiations distributed in their different frequencies or wavelengths. So, if we go from the smallest wavelengths known up to now (because according to physics the electromagnetic spectrum is infinite and continuous) to the longest, the electromagnetic spectrum covers the following radiations:  

Gamma rays, X-rays, ultraviolet, visible light, infrared, radio waves and microwaves:

-Gamma rays: With a wavelength in the order of 10^{-12}m, is a type of ionizing radiation capable of penetrating matter quite deeply and is able to cause serious damage to the nucleus of the cells.

-X rays: With a wavelength between 1m and 10km. It is invisible to the human eye, capable of crossing opaque bodies and is also an ionizing radiation.

-Ultraviolet light (UV): Whose wavelength is approximately between 100 nm and 380 nm; is a type of electromagnetic radiation that is not visible to the human eye.  

-Visible light: This part of the spectrum is located between ultraviolet light and infrared light (400 nm - 800 nm). It should be noted, the fact the only part of the whole electromagnetic spectrum is visible to humans is because the receptors in our eyes are only sensitive to these wavelengths.

-Infrared: This type of radiation is not visible to the human eye, since its wavelengths are outside the visible spectrum (between 700 nm and 1 mm).  

These waves can be divided into:  

- Near infrared or long wave infrared: it is the least sensitive to color and is easily absorbed by water.  

- Medium or medium wave infrared: it is also insensitive to color and easily absorbed by water and many types of plastics and paints.  

- Far infrared or short wave infrared: it is more penetrating than the long wave and is good for heating metals, these waves also can pass through clear materials.  

This light has many uses, including heating lamps in physiotherapy and medical treatments, heat sensing devices, among others.

-Radio waves: This is a type of electromagnetic radiation with wavelengths between 10 m to 10,000 m. This type of electromagnetic waves is very well reflected in the ionosphere, the layer of the atmosphere through which they travel directly or using repeaters. In addition, this radiation is important in telecommunications.

3 0
3 years ago
An electric immersion heater is put at the bottom of a large tank of water. The water next to the heater becomes warm
djyliett [7]

Answer:

when the water is heated with immersion heater, the water becomes less dense due to which the warm water rises up and the cooler water fills it's space.

4 0
2 years ago
According to Bode’s Law a planet is missing between Jupiter and Saturn. True False
defon
Bode law, a planet<span> was believed to exist </span>between<span> .... An Astronomer's Account of the </span>Missing Planet Between<span> Mars and </span>Jupiter<span> as Interpreted </span>Jupiter<span> ·</span>Saturn<span> · Uranus · Neptune.</span>
6 0
3 years ago
Three charges lie along the x axis. The positive charge q1 = 15 μC is at x= 2.0 m and the positive charge q2 = 6.0μC is at the o
Delicious77 [7]

Answer:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

F_{23} = \frac{k*q_{2}*q_{3}}{x^2} (Electric force of q₂ over q₃)

F_{13} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (Electric force of q₁ over q₃)

Known data

q₁ = 15 μC = 15*10⁻⁶ C

q₂ = 6 μC = 6*10⁻⁶ C

Problem development

F₂₃ = F₁₃

\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (We cancel k and q₃)

\frac{q_2}{x^2}=\frac{q_1}{(2-x)^2}

q₂(2-x)² = q₁x²

6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)

6(2-x)² = 15(x)²

6(4-4x+x²) = 15x²

24 - 24x + 6x² = 15x²

9x² + 24x - 24 = 0

The solution of the quadratic equation is:

x₁ = 0.775m

x₂ = -3.44m

x₁ meets the conditions for the forces to cancel in q₃

x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel

The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.

6 0
3 years ago
Read 2 more answers
A bus starts from rest.if the acceleration is 2m/s square, find
MrMuchimi

Answer:

The velocity after 2 seconds can be found through:

V = u +a*t

Where V is final velocity, u is initial velocity, a is acceleration and t is time.

V = 0 + 2* 2= 4 meters/second

The distance (s) can be found through:

V^2= u^2 +2*a* s

Where V is final velocity, u is initial velocity, a is acceleration.

4^2= 0^2 + 2 *2*s

16= 0 + 4s

s= 4 meters

Distance (s) can also be found through:

s= ut + 1/2 at^2

s= 0+ 1/2 *2*2^2= 1 *2*2

s= 4 meters

Explanation:

3 0
1 year ago
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