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3 years ago

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What is cutoff wavelength?

1 answer:

LenKa [72]3 years ago

3 0

Answer:

The cutoff wavelength is the minimum wavelength in which a particular fiber still acts as a single mode fiber. Above the cutoff wavelength, the fiber will only allow the LP01 mode to propagate through the fiber (fiber is a single mode fiber at this wavelength).

Explanation:

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Please help ! will mark brainliest < 3

love history [14]

Answer: 50m

Explanation:

Velocity is the Amount of Displacement an object makes, as Speed is the amount of Distance. As the problem is asking for displacement, find out how much he/she went in a straight line. He went from 10 - 60, so suptract, and get 50m.

4 0

3 years ago

At what angle two forces P + Q and (P - Q) act so that their resultant is :

stiv31 [10]

Use resultant formula

$\boxed{\sf R=\sqrt{A^2+B^2+2ABcos\theta}}$

So

#1

A be p+q and B be p-q

$\\ \rm\Rrightarrow R=\sqrt{3p^2+q^2}$

$\\ \rm\Rrightarrow \sqrt{(p+q)^2+(p-q)^2+2(p+q)(p-q)cos\alpha}=\sqrt{3p^2+q^2}$

$\\ \rm\Rrightarrow 2p^2+2q^2+2(p^2-q^2)cos\alpha=3p^2+q^2$

$\\ \rm\Rrightarrow 2cos\alpha=1$

$\\ \rm\Rrightarrow cos\alpha=\dfrac{1}{2}$

$\\ \rm\Rrightarrow \alpha=\dfrac{\pi}{3}$

#2

$\\ \rm\Rrightarrow 2(p^2+q^2)+2(p^2-q^2)cos\beta=2(p^2+q^2)$

$\\ \rm\Rrightarrow 2cos\beta=0$

$\\ \rm\Rrightarrow cos\beta=0$

$\\ \rm\Rrightarrow \beta=\dfrac{\pi}{2}$

#3

$\\ \rm\Rrightarrow 2(p^2+q^2)+2(p^2-q^2)cos\gamma=p^2+q^2$

$\\ \rm\Rrightarrow 2(p^2-q^2)cos\gamma=-(p^2+q^2)$

$\\ \rm\Rrightarrow cos\gamma =\dfrac{q^2-p^2}{2(p^2-q^2)}$

$\\ \rm\Rrightarrow \gamma=cos^{-1}\left(\dfrac{q^2-p^2}{2(p^2+q^2)}\right)$

8 0

2 years ago

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Calculate the average speed of a gazelle that runs 140 meters in 5.0 seconds.

atroni [7]

<span>The average speed is 14 m / sec or 50.4 km / h.</span>

8 0

3 years ago

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A leaky 10-kg bucket is lifted from the ground to a height of 10 m at a constant speed with a rope that weighs 0.9 kg/m. Initial

sladkih [1.3K]

Answer:

The work done shall be 14715 Joules

Explanation:

The work done by a force 'F' in a displacement 'dy' is given by

$W=m(y)g\times dy$

At any position 'y' the weight shall be sum of weft of water and weight of string

$\therefore m(y)=m_{water}(y)+m_{string}(y)\\\\m(y)=30(1-\frac{y}{10})+0.9y$

Thus applying values we get

$W=\int m(y)g\times dy\\\\W=\int_{0}^{10}(30(1-\frac{y}{10}+0.9y)\times g)dy\\\\Solving\\W=294.3\int_{0}^{10}(1-\frac{y}{10}+0.9y)dy\\\\W=14715J$

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2 years ago

A bicycle rider travels 50.0 Km in 2.5 hours. What is the cyclist's average speed?

vekshin1

<span>If you divide the numbers you will get on average of 20km</span>

5 0

3 years ago

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