What are star system

Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light.

pogonyaev

Complete Question

Due to blurring caused by atmospheric distortion, the best resolution that can be obtained by a normal, earth-based, visible-light telescope is about 0.3 arcsecond (there are 60 arcminutes in a degree and 60 arcseconds in an arcminute).Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light

Answer:

The diameter is $D = 0.59 \ m$

Explanation:

From the question we are told that

The best resolution is $\theta = 0.3 \ arcsecond$

The wavelength is $\lambda = 700 \ nm = 700 *10^{-9 } \ m$

Generally the

1 arcminute = > 60 arcseconds

=> x arcminute => 0.3 arcsecond

So

$x = \frac{0.3}{60 }$

=> $x = 0.005 \ arcminutes$

Now

60 arcminutes => 1 degree

0.005 arcminutes = > z degrees

=> $z = \frac{0.005}{60 }$

=> $z = 8.333 *10^{-5} \ degree$

Converting to radian

$\theta = z = 8.333 *10^{-5} * 0.01745 = 1.454 *10^{-6} \ radian$

Generally the resolution is mathematically represented as

$\theta = \frac{1.22 * \lambda }{ D}$

=> $D = \frac{1.22 * \lambda }{\theta }$

=> $D = \frac{1.22 * 700 *10^{-9} }{ 1.454 *10^{-6} }$

=> $D = 0.59 \ m$

4 0

3 years ago

__________ is when a person becomes less sensitive to a drug. A. Tolerance B. Craving C. Addiction D. Dependence

Luden [163]

A) Tolerance

Tolerance is developed after using a drug repeatedly, so the body adapts to it. Because of that, people who develop a tolerance would then need to use more of that drug to get the same effect.

4 0

3 years ago

A car travels at a steady 40.0 m/s around a horizontal curve of radius 200 m. What is the minimumcoefficient of static friction

zhenek [66]

Answer:

c. 0.816

Explanation:

Let the mass of car be 'm' and coefficient of static friction be 'μ'.

Given:

Speed of the car (v) = 40.0 m/s

Radius of the curve (R) = 200 m

As the car is making a circular turn, the force acting on it is centripetal force which is given as:

Centripetal force is, $F_c=\frac{mv^2}{R}$

The frictional force is given as:

Friction = Normal force × Coefficient of static friction

$f=\mu N$

As there is no vertical motion, therefore, $N=mg$ . So,

$f=\mu mg$

Now, the centripetal force is provided by the frictional force. Therefore,

Frictional force = Centripetal force

$f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu=\frac{v^2}{Rg}$

Plug in the given values and solve for 'μ'. This gives,

$\mu = \frac{(40\ m/s)^2}{200\ m\times 9.8\ m/s^2}\\\\\mu=\frac{1600\ m^2/s^2}{1960\ m^2/s^2}\\\\\mu=0.816$

Therefore, option (c) is correct.

7 0

3 years ago

Five identical quintuplets leave earth when they reach the age of 21, in the year 2121. Each quintuplet goes on a spaceship jour

Elena-2011 [213]

Answer:

Explanation:

This is a problem based on time dilation , a theory given by Albert Einstein .

The formula of time dilation is as follows .

t₁ = $\frac{t}{\sqrt{1-\frac{v^2}{c^2} } }$

t is time measured on the earth and t₁ is time measured by man on ship .

A ) Given t = 20 years , t₁ = ? v = .4c

$\frac{20}{\sqrt{1-\frac{.16c^2}{c^2} } }$

=1.09 x 20

t₁= 21.82 years

B ) Given t = 5 years , t₁ = ? v = .2c

$\frac{5}{\sqrt{1-\frac{.04c^2}{c^2} } }$

=1.02 x 5

t₁= 5.1 years

C ) Given t = 10 years , t₁ = ? v = .8c

$\frac{10}{\sqrt{1-\frac{.64c^2}{c^2} } }$

=1.67 x 10

t₁= 16.7 years

D ) Given t = 10 years , t₁ = ? v = .4c

$\frac{10}{\sqrt{1-\frac{.16c^2}{c^2} } }$

=1.09 x 10

t₁= 10.9 years

E ) Given t = 20 years , t₁ = ? v = .8c

$\frac{20}{\sqrt{1-\frac{.64c^2}{c^2} } }$

=1.67 x 20

t₁= 33.4 years

7 0

3 years ago

In the flow past a compression corner, the upstream Mach number and pressure are 3.5 and 1 atm, respectively. Downstream of the

yulyashka [42]

Answer:

$\theta=23.7^{\circ}$

Explanation:

The ratio of pressure 2 to 1 us 5.48/1= 5.48 rounded off as 5.5.

Referring to table A.2 of modern compressible flow then $M_{\beta_1}=2.2$

Also

$M_{\beta_1}=M_1 sin \beta$ and making $sin\beta$ the subject of the formula then

$sin\beta=\frac {2.2}{3.5}\\\beta=38.94^{\circ}$

Making reference to $\theta-\beta-M$ diagram then

$\theta=23.7^{\circ}$

4 0

3 years ago