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sergiy2304 [10]
3 years ago
9

Speed of light in air is 3 x 10 m/s, determine the speed of light in the glass fibre.

Physics
1 answer:
Dafna11 [192]3 years ago
4 0
30 speed of light in the glass
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This question doesn't appear to be complete
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3 years ago
Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. The electric field at a po
gregori [183]

Answer:

E = 0    r <R₁

Explanation:

If we use Gauss's law

      Ф = ∫ E. dA = q_{int} / ε₀

in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.

Consequently by Gauss's law the electric field is ZERO

           E = 0    r <R₁

6 0
3 years ago
What control what enter and leaves the cell in a animal cell
Rom4ik [11]
The Cell Membrane is what controls what enters and leaves the cell.
5 0
3 years ago
A magnet falls through a loop of wire with the south pole entering first. After it has fallen all the way through the wire loop
Alisiya [41]

Answer:

The induced current direction as viewed is clockwise

Explanation:

Lenz's Law states that the induced e. m. f. causes current to be driven in the loop of wire in such a way as to generate magnetic field that are oppose the magnetic flux change which is the source of the induced current

Therefore, as the magnet approaches the coil with the south pole,  the coil produces current equivalent to the upward movement of the south pole of a permanent magnet through it which according to Flemings Right Hand Rule is clockwise

Therefore;

The direction of the induced current in the loop (as viewed from above, looking down the magnet) is clockwise

5 0
3 years ago
The standard emf for the cell using the overall cell reaction below is +2.20 V: 2Al(s) + 3I2 (s) → 2Al3+ (aq) + 6I− (aq) The emf
Triss [41]

Answer: +2.10V

Explanation:

2Al(s)+3I_2(s)\rightarrow 2Al^{3+}(aq)+6I^-(aq)

Using Nernst equation :

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log K

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log [Al^{3+}]^2\times [I^-]^6

where,

E^o_{cell} = standard emf for the cell = +2.20 V

n = number of electrons in oxidation-reduction reaction = 6

E_{cell} = emf of the cell = ?

[Al^{3+}]  = concentration = 5.0\times 10^{-3}M

[I}^{-}]  = concentration = 0.10M

Now put all the given values in the above equation, we get:

E_{cell}=+2.20-\frac{0.059}{6}\log [5.0\times 10^{-3}]^2\times [0.10]^6

E_{cell}=2.10V

The standard emf for the cell using the overall cell reaction below is +2.10 V

5 0
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