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

=> 
=>
=>
Explanation :
It is given that,
BMR i.e basal metabolic rate is 88 kcal/hr. So, BMR in watts is converted by the following :
We know that, 1 kilocalorie = 4184 joules
So, 

J/sec is nothing but watts.
So, 
and 
So, it can be seen that the body can accommodate a modes amount of activity in hot weather but strenuous activity would increase the metabolic rate above the body's ability to remove heat.
The total work done by the electric field on the charge is given by the scalar product between the electric force acting on the charge and the displacement of the charge:

where the force is F=qE, d=0.556 and

. Using the value of q and E given by the problem, we find
Answer:
You will hear the note E₆
Explanation:
We know that:
Your speed = 88m/s
Original frequency = 1,046 Hz
Sound speed = 340 m/s
The Doppler effect says that:

Where:
f = original frequency
f' = new frequency
v = velocity of the sound wave
v0 = your velocity
vs = velocity of the source, in this case, the source is the diva, we assume that she does not move, so vs = 0.
Replacing the values that we know in the equation we have:

This frequency is close to the note E₆ (1,318.5 Hz)
Answer:
By plotting the locations of earthquakes
Explanation:
When plotting the locations of earthquakes, scientists have been able to locate plate boundaries and also be able to determine plate characteristics and predict the movement of plates