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3 years ago

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The cross sectional area of an optical fibre is 6.3 x 10^-6 m^2 The intensity of the light entering the optical fibre is 3.2 x 1

0^7 W/m^2 Calculate the power of the light entering the optical fibre

1 answer:

gogolik [260]3 years ago

7 0

the answer is 0.0. also known as a emoji face!

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Someone please help !!

masha68 [24]

I’m pretty sure it’s c.... hope it helps and hope it’s right.

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3 years ago

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

b

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$

$\frac{m(a+b)^2}{3} \omega_3^2 + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0$

$\frac{1.53(0.6+0.6)^2}{3} \omega_3^2 + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0$

$0.7344 \omega_3^2 = 2.128$

$\omega _3 = \sqrt{\frac{2.128}{0.7344} }$

$\omega _3 =1.70 \ rad/s$

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

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3 years ago

HELP ME HOW ARE BLACK HOLES FROMED

V125BC [204]

Answer:

Stellar black holes form when the center of a very massive star collapses in upon itself.

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2 years ago

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15) Very cold climates occur at Earth's North and South Poles because the polar regions

Gwar [14]

Because the polar regions receive low-angle insolation.

Insolation is the amount of solar radiation received by a given area. The Sun is always low on the horizon. The low Sun angle makes the beam of solar radiation to travel a longer distance from upper troposphere to reach earth's surface as compared to when it is directly overhead. In this case, the radiations are scattered and reflected more by the atmosphere and spread over a larger area. Thus, the intensity of solar radiation is very less at polar regions than near the equatorial region. This is the reason of very cold climates at polar regions.

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2 years ago

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In order to waken a sleeping child, the volume on an alarm clock is doubled. Under this new scenario, how much more energy will

Kamila [148]

Answer:4 times more energy will be striking the childbearing

Explanation:

Because Volume is directly proportional to amplitude of sound. Energy is proportional to amplitude squared. If you triple the amplitude, you multiply the energy by 4

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2 years ago