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 
Explanation:
From the question we are told that
The best resolution is 
The wavelength is 
Generally the
1 arcminute = > 60 arcseconds
=> x arcminute => 0.3 arcsecond
So
=> 
Now
60 arcminutes => 1 degree
0.005 arcminutes = > z degrees
=> 
=> 
Converting to radian
Generally the resolution is mathematically represented as
=> 
=> 
=>
Answer:
A) a = 2.31[m/s^2]; B) t = 14.4 [s]
Explanation:
We can solve this problem using the kinematic equations, but firts we must identify the data:
Vf= final velocity = take off velocity = 120[km/h]
Vi= initial velocity = 0, because the plane starts to move from the rest.
dx= distance to run = 240 [m]
![v_{f} ^{2} =v_{i} ^{2}+2*g*dx\\where:\\v_{f}=120[\frac{km}{h} ]*\frac{1hr}{3600sg} * \frac{1000m}{1km} =33.33[m/s]\\\\Replacing\\33.33^{2}=0+2*a*(240)\\ a=\frac{11108.88}{2*240}\\ a=2.31[m/s^2]\\](https://tex.z-dn.net/?f=v_%7Bf%7D%20%5E%7B2%7D%20%3Dv_%7Bi%7D%20%5E%7B2%7D%2B2%2Ag%2Adx%5C%5Cwhere%3A%5C%5Cv_%7Bf%7D%3D120%5B%5Cfrac%7Bkm%7D%7Bh%7D%20%5D%2A%5Cfrac%7B1hr%7D%7B3600sg%7D%20%2A%20%5Cfrac%7B1000m%7D%7B1km%7D%20%3D33.33%5Bm%2Fs%5D%5C%5C%5C%5CReplacing%5C%5C33.33%5E%7B2%7D%3D0%2B2%2Aa%2A%28240%29%5C%5C%20a%3D%5Cfrac%7B11108.88%7D%7B2%2A240%7D%5C%5C%20%20a%3D2.31%5Bm%2Fs%5E2%5D%5C%5C)
To find the time we must use another kinematic equation.
![v_{f} =v_{i} +a*t\\replacing:\\33.33=0+(2.31*t)\\t=\frac{33.33}{2.31}\\ t=14.4[s]](https://tex.z-dn.net/?f=v_%7Bf%7D%20%3Dv_%7Bi%7D%20%2Ba%2At%5C%5Creplacing%3A%5C%5C33.33%3D0%2B%282.31%2At%29%5C%5Ct%3D%5Cfrac%7B33.33%7D%7B2.31%7D%5C%5C%20t%3D14.4%5Bs%5D)
I'm pretty sure this is A
Answer:
0.29D
Explanation:
Given that
F = G M m / r2
F = GM(6m) / (D-r)2
G Mm/r2 = GM(6m) / (D-r)2
1/r2 = 6 / (D-r)2
r = D / (Ö6 + 1)
r = 0.29 D
See diagram in attached file
For a pendulum with a massless rope 'L' meters long, swinging through
a small arc, the period of the swing is
2 π √(L/g) seconds .
and it doesn't depend on the mass of the thing on the end of the rope ...
that could be a pebble or a bus.
For the girl on the 2.5-m rope,
Period = (2 π) √(2.5 / 9.8) = 3.17 seconds
Frequency = 1 / period = about <em> 0.315 Hz .</em>
