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3 years ago

Which type of telescope focuses star light using mirrors?

1 answer:

7 0

Coudé telescopes use a convex secondary mirror like a Cassegrain and an angled mirror like a Newtonian reflector to move the light rays to a focal point away from the telescope. This arrangement is useful when optical equipment is being used that is too heavy to mount directly on the telescope.

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A beam of light from a laser illuminates a glass how long will a short pulse of light beam take to travel the length of the glas

Eduardwww [97]

Answer:

The time of short pulse of light beam is $2.37\times10^{-9}\ sec$

Explanation:

Given that,

A beam of light from a laser illuminates a glass.

Suppose, the length of piece is $L=25.21\times10^{-2}\ m$

Index of refraction is 2.83.

We need to calculate the speed of light pulse in glass

Using formula of speed

$v=\dfrac{c}{\mu}$

Put the value into the formula

$v=\dfrac{3\times10^{8}}{2.83}$

$v=1.06\times10^{8}\ m/s$

We need to calculate the time of short pulse of light beam

Using formula of velocity

$v=\dfrac{d}{t}$

$t=\dfrac{d}{v}$

Put the value into the formula

$t=\dfrac{25.21\times10^{-2}}{1.06\times10^{8}}$

$t=2.37\times10^{-9}\ sec$

Hence, The time of short pulse of light beam is $2.37\times10^{-9}\ sec$

3 0

4 years ago

I need help with this please

34kurt

The answer is Rubber

5 0

3 years ago

If a display has a dynamic range of 20 dB and the smallest voltage it can handle is 200 mV, then the largest voltage it can hand

DedPeter [7]

Answer:

The largest voltage is 0.02 V.

(d) is correct option.

Explanation:

Given that,

Range = 20 dB

Smallest voltage = 200 mV

We need to calculate the largest voltage

Using formula of voltage gain

$G_{dB}=10 log_{10}(\dfrac{V_{out}^2}{V_{in}^{2}})$

$20 =10 log_{10}(\dfrac{V_{out}^2}{V_{in}^{2}})$

$2=log_{10}(\dfrac{V_{out}^2}{V_{in}^{2}})$

$10^2=\dfrac{V_{out}^2}{V_{in}^{2}}$

$\dfrac{V_{out}}{V_{in}^}=10$

$V_{in}=\dfrac{V_{out}}{10}$

$V_{in}=\dfrac{200}{10}$

$V_{in}=20\ mV$

$V_{in}=0.02\ V$

Hence, The largest voltage is 0.02 V.

8 0

3 years ago

A particle's position along the x-axis is described by. x(t)= At+Bt^2where t is In seconds: x is in meters: and the constants A

DanielleElmas [232]

Answer

given,

position of particle

x(t)= A t + B t²

A = -3.5 m/s

B = 3.9 m/s²

t = 3 s

a) x(t)= -3.5 t + 3.9 t²

velocity of the particle is equal to the differentiation of position w.r.t. time.

$\dfrac{dx}{dt}=\dfrac{d}{dt}(-3.5t + 3.9t^2)$

$v= -3.5 + 7.8 t$ ------(1)

velocity of the particle at t = 3 s

v = -3.5 + 7.8 x 3

v = 19.9 m/s

b) velocity of the particle at origin

time at which particle is at origin

x(t)= -3.5 t + 3.9 t²

0 = t (-3.5 + 3.9 t )

t = 0, $t=\dfrac{3.5}{3.9}$

t = 0 , 0.897 s

speed of the particle at t = 0.897 s

from equation (1)

v = -3.9 + 7.8 t

v = -3.9 + 7.8 x 0.897

v = 3.1 m/s

6 0

4 years ago

Can someone help me please? I am stuck.

Mars2501 [29]

The distance traveled and the time it takes to travel are both going up at a steady rate. The distance goes from lap 1, lap 2, lap 3, lap 4. The time is increasing by a minute each time. When both the x and y scenario are increasing at a STEADY RATE the line will be linear. B displays a linear line.

8 0

3 years ago