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lidiya [134]
3 years ago
6

PLZ!!!!!HURRY WILL BE FRIEND FOREVER!!!

Physics
1 answer:
Bad White [126]3 years ago
3 0

The reason why weight of ashes resulting from the burning of wood is not equal to the weight of wood piece before it burns is explained below

<u>Explanation:</u>

  • The reason why the weight of ashes resulting from the burning of wood is not equal to the wood piece before it burns is that when wood burns it loses its mass and that's how ashes are created.
  • Wood contains carbon and hydrogen when these two combines with oxygen at high temperatures Carbon dioxide and water vapor are obtained as by product.when this mixture leaves the wood in the form of smoke only burnt ashes are leftover.
  • This is the main reason for inequality in the weight of wood and ash after burning.
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Think about the way optical fibers are designed. why are the ends of the fibers glowing very brightly while the sides of the fib
Effectus [21]

Answer:

D. because the light is reflected back into the fiber along its sides

Explanation:

The fiber is constructed in a way that the light is bent/reflected/refracted toward the center core of glass. So, from the center core, there is a layer above it that has a different propagation than the core, and above that the same thing. To give you a real world visual example, if you look down in a pool of water, then stick a straight stick into it, you see that the straight stick appears to bend. That is what is happening to the light as it travels through a different medium (air to water). This same effect is incorporated in the fiber optic cable construction.

6 0
2 years ago
The first law of thermodynamics states that energy can neither be created nor destroyed. If this is true then why are we always
riadik2000 [5.3K]

Answer:

The second law of thermodynamics states in an isolated system, the entropy (the amount of thermal energy that cannot be converted into mechanical work, also known as the amount of disorder) always increases, therefore, an isolated system always require an external input (new sources) of energy for there to be orderliness or for the available energy of the system to remain constant or increase

Explanation:

5 0
3 years ago
What is #6<br><br> IM GIVING 40 POINTS
frosja888 [35]

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4 0
3 years ago
A charge of -2.65 nC is placed at the origin of an xy-coordinate system, and a charge of 2.00 nC is placed on the y axis at y =
stiks02 [169]

Answer:

A. Fnx = 5.71*10⁻⁵ N  ,  Fny= -3.67*10⁻⁵ N

B. Fn= 6.78 *10⁻⁵ N

C. α= 32.4° counterclockwise with the positive x+ axis

Explanation:

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Equivalences

1nC= 10⁻⁹C

1cm = 10⁻²m

Known data

k= 9*10⁹N*m²/C²

q₁= -2.65 nC =-2.65*10⁻⁹C

q₂= +2.00 nC = 2*10⁻⁹C

q₃= +5.00 nC= =+5*10⁻⁹C

d_{13} = \sqrt{(3.2)^{2} +(3.8)^{2} }

d_{13} =\sqrt{24.68} * 10⁻²m    = 4.9678* 10⁻²m

(d₁₃)² = 24.68*10⁻⁴m²

d₂₃ = 3.2 cm = 3.2*10⁻²m  

Graphic attached

The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.

The force (F₂₃) of q₂ on q₃ is repulsive because the charges have equal signs and the forces.

The force (F₁₃) of q₁ on q₃ is attractive because the charges have opposite signs.

Magnitudes of F₁₃ and F₂₃

F₁₃ = (k*q₁*q₃)/(d₁₃)²=( 9*10⁹*2.65*10⁻⁹*5*10⁻⁹) /(24.68*10⁻⁴)

F₁₃ = 4.8 *10⁻⁵ N

F₂₃ = (k*q₂*q₃)/(d₂₃)² =  ( 9*10⁹*2*10⁻⁹*5*10⁻⁹) /((3.2)²*10⁻⁴)

F₂₃ = 8.8 *10⁻⁵ N

x-y components of F₁₃ and F₂₃

F₁₃x= -4.8 *10⁻⁵ *cos β= - 4.8 *10⁻⁵(3.2/ (4.9678)= - 3.09*10⁻⁵ N

F₁₃y= -4.8 *10⁻⁵ *sin β= - 4.8 *10⁻⁵(3.8/(4.9678) =  - 3.67*10⁻⁵ N

F₂₃x  = F₂₃ =  +8.8 *10⁻⁵ N

F₂₃y = 0

x and y components of the total force exerted on q₃ by q₁ and q₂ (Fn)

Fnx= F₁₃x+F₂₃x =  - 3.09*10⁻⁵ N+8.8 *10⁻⁵ N= 5.71*10⁻⁵ N

Fny= F₁₃y+F₂₃y = - 3.67*10⁻⁵ N+0= - 3.67*10⁻⁵ N

Fn magnitude

F_{n} =\sqrt{(Fn_{x})^{2}+(Fn_{y})^{2}  }

F_{n} = \sqrt{(5.71)^{2}+(3.67)^{2}  } *10⁻⁵ N

Fn= 6.78 *10⁻⁵ N

Fn direction  (α)

\alpha =tan^{-1}( \frac{Fn_{y} }{Fn_{x} } )

\alpha =tan^{-1}( \frac{-3.67 }{5.71} )

α= -32.4°

α= 32.4° counterclockwise with the positive x+ axis

4 0
3 years ago
What is the resistance of a 3.5 m copper wire (Rho= 1.7x10-8 Ohm·m) that 1 point
VikaD [51]

Answer:

(D)

Explanation:

Given :

l=3.5 m

A=5.26*10^{-6} m^{2}

p=1.7*10^{-8}  ohm.m

Resistance can be calculated as :

R=p\frac{l}{A} \\R=1.7*10^{-8} \frac{3.5}{5.26*10^{-6} }

R=\frac{5.95*10^{-2} }{5.26} \\R=1.13*10^{-2}

Resistance of the wire will be 1.1×10^{-2} ohms

Option D is correct

4 0
2 years ago
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