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Stella [2.4K]
4 years ago
15

PLEASE I NEED HELP ASAPPPP

Physics
2 answers:
lesya692 [45]4 years ago
7 0


1. 4

2. carbon

3. Bond

blondinia [14]4 years ago
5 0
1. Each carbon atom has 4 valence electrons for forming bonds. 
2. C<span>arbon atoms can bond with carbon atoms and with other elements in many different ways. 
3. A(n) bond is the force that holds two atoms together. 
Hope I helped, and good luck! (:</span>
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A length of copper wire carries a current of 14 A, uniformly distributed through its cross section. The wire diameter is 2.5 mm,
gavmur [86]

Answer:

a. ρ_\beta=1.996J/m^3

b. U_E=9.445x10^{-15} J/m^3

Explanation:

a. To find the density of magnetic field given use the gauss law and the equation:

i=14A, d=2.5mm, R=3.3Ω, l=1 km, E_o=8.85x10^{-12}F/m, u_o=4*x10^{-7}H/m

ρ_\beta=\frac{\beta^2}{2*u_o}

ρ_\beta=\frac{1}{2*u_o}*(\frac{u_o*i^2}{2\pi *r})^2

ρ_\beta=\frac{u_o*i^2}{8\pi*r}=\frac{4\pi *10^{-7}H/m*(14A)^2}{8\pi*(1.25x10^{-3}m)^2}

ρ_\beta=1.996J/m^3

b. The electric field can be find using the equation:

U_E=\frac{1}{2}*E_o*E^2

E=(\frac{i*R}{l})^2

U_E=\frac{1}{2}*8.85x10^{-12}*(\frac{14A*3.3}{1000m})^2

U_E=9.445x10^{-15} J/m^3

4 0
3 years ago
Within the theory of G relativity what, exactly, is meant by " the speed of light WITHIN A VACUUM" ? &amp; what does that have t
Ber [7]
The speed of light "within a vacuum" refers to the speed of electromagnetic radiation propagating in empty space, in the complete absence of matter.  This is an important distinction because light travels slower in material media and the theory of relativity is concerned with the speed only in vacuum.  In fact, the theory of relativity and the "speed of light" actually have nothing to do with light at all.  The theory deals primarily with the relation between space and time and weaves them into an overarching structure called spacetime.  So where does the "speed of light" fit into this?  It turns out that in order to talk about space and time as different components of the same thing (spacetime) they must have the same units.  That is, to get space (meters) and time (seconds) into similar units, there has to be a conversion factor.  This turns out to be a velocity.  Note that multiplying time by a velocity gives a unit conversion of
seconds \times  \frac{meters}{seconds} =meters
This is why we can talk about lightyears.  It's not a unit of time, but distance light travels in a year.  We are now free to define distance as a unit of time because we have a way to convert them.  
As it turns out light is not special in that it gets to travel faster than anything else.  Firstly, other things travel that fast too (gravity and information to name two).  But NO events or information can travel faster than this.  Not because they are not allowed to beat light to the finish line---remember my claim that light has nothing to do with it.  It's because this speed (called "c") converts space and time.  A speed greater than c isn't unobtainable---it simply does not exist.  Period.  Just like I can't travel 10 meters without actually moving 10 meters, I cannot travel 10 meters without also "traveling" at least about 33 nanoseconds (about the time it takes light to get 10 meters)  There is simply no way to get there in less time, anymore than there is a way to walk 10 meters by only walking 5.  
We don't see this in our daily life because it is not obvious that space and time are intertwined this way.  This is a result of our lives spent at such slow speeds relative to the things around us.
This is the fundamental part to the Special Theory of Relativity (what you called the "FIRST" part of the theory)  Here is where Einstein laid out the idea of spacetime and the idea that events (information) itself propagates at a fixed speed that, unlike light, does not slow down in any medium.  The idea that what is happening "now" for you is not the same thing as what is "now" for distant observers or observers that are moving relative to you.  It's also where he proposed of a conversion factor between space and time, which turned out to be the speed of light in vacuum.
3 0
4 years ago
mass of the planet is 12 times that of earth and its radius is thrice that of earth , then find the escape velocity on that plan
Over [174]

Answer:

The escape velocity on the planet is approximately 178.976 km/s

Explanation:

The escape velocity for Earth is therefore given as follows

The formula for escape velocity, v_e, for the planet is v_e = \sqrt{\dfrac{2 \cdot G \cdot m}{r} }

Where;

v_e = The escape velocity on the planet

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m = The mass of the planet = 12 × The mass of Earth, M_E

r = The radius of the planet = 3 × The radius of Earth, R_E

The escape velocity for Earth, v_e_E, is therefore given as follows;

v_e_E = \sqrt{\dfrac{2 \cdot G \cdot M_E}{R_E} }

\therefore v_e = \sqrt{\dfrac{2 \times G \times 12 \times M}{3 \times R} } =  \sqrt{\dfrac{2 \times G \times 4 \times M}{R} } = 16 \times \sqrt{\dfrac{2 \times G \times M}{R} } = 16 \times v_e_E

v_e = 16 × v_e_E

Given that the escape velocity for Earth, v_e_E ≈ 11,186 m/s, we have;

The escape velocity on the planet = v_e ≈ 16 × 11,186 ≈ 178976 m/s ≈ 178.976 km/s.

3 0
3 years ago
A reaction mixture in a 3.67 L flask at a certain temperature initially contains 0.763 g H2 and 96.9 g I2, At equilibrium, the f
Alexxandr [17]

Answer:

13 530 482

Explanation:

                            H2    +          I2     ------>      2HI

start (mol)             0.3785         0.3818                   0

change (mol)       -0.3534        -0.3534            +0.7067

equilibrium (mol)  0.0251         0.0284             0.7067

concentra (mol/L) 0.0068        0.0077              0.1926

K_{c} = \frac{0.1926^{2}}{0.0068^{2}*0.0077^{2} } = 13530482

7 0
3 years ago
Enunciado: Una bola se lanza verticalmente de la parte superior de un edificio con una velocidad inicial de 25 m/s. La bola impa
bearhunter [10]

Responder:

Explicación:

Usaremos la ecuación de movimiento para determinar la altura de la bola medida desde la parte superior del edificio.

Usando la ecuación para obtener la altura de caída

S = ut + 1 / 2gt²

u es la velocidad inicial = 25 m / s

g es la aceleración debida a la gravedad = 9,81 m / s²

t es el tiempo = 7 segundos

S es la altura de la caída

S = 25 (7) +1/2 (9,81) × 7²

S = 175 + 4,905 (49)

S = 175 + 240,345

S = 415,35 m

Esto significa que la pelota se elevó a 415,35 m de altura

7 0
3 years ago
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