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Evgesh-ka [11]
2 years ago
15

In the design of an optical fibre, what

Physics
2 answers:
MakcuM [25]2 years ago
7 0

Answer:

silica glass

Explanation:

denpristay [2]2 years ago
4 0

Hey there! I'll try to provide you with my best answer.

Answer: pure glass (SiO2), plastic, or a combination of both is most suitable for the  design of the core. Even so in cladding. The use of one or the other material will be determined by such factors as quality and economics.

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A boy standing throws a penny horizontally at 7.25 m/s out of the window of his apparent buliding. If the window is 10.0 m above
Ksivusya [100]

Answer:

Explanation:

This is a 2D problem (parabolic) so we have to think that way. We have to split up the problem into its 2 dimensions to solve it. Think "y-stuff" and "x-stuff".

In the y-stuff category:

v₀ = 0 (initial upwards velocity is 0 since we are told the penny is thrown horizontally)

Δx = -10.0 m (this displacement is negative because the penny lands 10.0 m below the point from which it was thrown)

a = -9.8 m/s/s

t = ? (we need to find the time in this dimension so we can use it in the x dimension to find the displacement, our unknown)

In the "x-stuff" category:

v₀ = 7.25 m/s (this is given)

Δx = ???

a = 0 (acceleration in this dimension is ALWAYS 0)

t = (we will solve for this in the y-dimension and plug it in here).

In the y dimension:

Δx = v₀t + \frac{1}{2}at^2 and plugging in from the y-dimension info:

-10.0=0t+\frac{1}{2}(-9.8)t^2 which simplifies to

-10.0=-4.9t^2 so

t=\sqrt{\frac{-10.0}{-4.9} } which, to 2 significant digits is

t = 1.4 seconds

Now we will do the same in the x-dimension, using t = 1.4:

Δx = v₀t + \frac{1}{2}at^2 and filling in the x-stuff:

Δx = 7.25(1.4)+\frac{1}{2}(0)(1.4)^2 Notice that the stuff after the + sign goes to 0 cuz of the multiplication of 0, so what we are left with is another form of the d = rt equation:

Δx = 7.25(1.4) + 0 so

Δx = 1.0 × 10¹ m (That's rounded correctly to 2 sig dig's: 10 m from the base of the building).

6 0
3 years ago
Read 2 more answers
A 4.0-cm tall light bulb is placed at distance of 8.3 cm from a concave mirror having a focal length of 15.2 cm. Determine the i
scoundrel [369]

Answer: the image distance is -18, 28 cm this means behind of the concave mirror. The image size is 2.2 higher that the original  so it has 8.8 cm with the same orientation as original  and it is a virtual imagen.

Explanation: In order to sove the imagen formation for a concave mirror we have to use the following equation:

1/p+1/q=1/f  where p and q represents the distance to the mirror  for the object and imagen, respectively. f is the focal length for the concave mirror.

replacing the values we obtain:

1/8.3+1/q=1/15.2

so 1/q=(1/15.2)-(1/8.3)=-54.7*10^-3

then q=-18.28 cm

The magnification is given by M=-q/p=-(-18,28)/8.3= 2.2

We also add a picture to see the imagen formation for this case.

6 0
3 years ago
Two protons are 9.261 fm apart. (1 fm= 1 femtometer = 1 x 10^-15 m.) What is the ratio of the electric force to the gravitationa
Strike441 [17]

Answer:

1.24\times 10^{36}

Explanation:

q = magnitude of charge on each proton = 1.6 x 10⁻¹⁹ C

m = mass of each proton = 1.67 x 10⁻²⁷ kg

r = distance between the two protons = 1 x 10⁻¹⁵ m

Electric force between the two protons is given as

F_{e} = \frac{kq^{2}}{r^{2}}

F_{e} = \frac{(9\times 10^{9})(1.6\times 10^{-19})^{2}}{(1\times 10^{-15})^{2}}

F_{e} = 230.4 N

Gravitational force between the two protons is given as

F_{g} = \frac{Gm^{2}}{r^{2}}

F_{g} = \frac{(6.67\times 10^{-11})(1.67\times 10^{-27})^{2}}{(1\times 10^{-15})^{2}}

F_{g} = 1.86\times 10^{-34} N

Ratio is given as

Ratio =\frac{F_{e}}{F_{g}}

Ratio =\frac{230.4}{1.86\times 10^{-34}}

Ratio = 1.24\times 10^{36}

3 0
3 years ago
Read 2 more answers
According to the video, what do Electrical Engineers usually study in college? Check all that apply.
Dmitry [639]

Answer:

Math & Science

Explanation:

You said so

6 0
3 years ago
Read 2 more answers
The pressure in a compressed air storage tank is 1200 kPa. What is the tank’s pressure in (a) kN and m units; (b) kg, m, and s u
elixir [45]

Explanation:

Given data

Pressure P=1200 kPa

To find

Pressure in

(a) kN/m²

(b) kg/m.s²

(c) kg/km.s²

Solution

For Part (a)

P=(1200kPa)(\frac{1kN/m^{2} }{1kPa} )\\P=1200kN/m^{2}

For Part (b)

P=(1200kPa)(\frac{1kN/m^{2} }{1kPa} )(\frac{1000kg.m/s^{2} }{1kN} )\\P=1,200,000kg/m.s^{2}

For Part (c)

P=(1200kPa)(\frac{1kN/m^{2} }{1kPa} )(\frac{1000kg.m/s^{2} }{1kN} )(\frac{100m}{1km} )\\P=1,200,000,000kg/km.s^{2}

8 0
3 years ago
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