<span>1. Why is photographing lightning a difficult process?
</span>Some reasons which come to my mind for saying this are: 1) You get only one chance for the particular situation - it is not like portrait photography where you can go back in the studio if the photos didn't come out well; 2) lightning varies so much in brightness, intensity and location that guessing the proper exposure requires a lot of experience, as well as luck; 3) you are always at some risk when photographing worthwhile lightning; and 4) lightning is a point (line) source, and demands the most of the optical quality of your camera
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2. What piece of equipment is helpful in capturing lightning photographs?
</span><span><span>SLR camera with B-shutter speed (preferably SLR; you might try using your digital camera, if it has B mode, but this is much more difficult)</span><span>lenses ranging from 28mm to 135mm at minimum. Fixed-focal lenses are preferred over zoomlenses. Aperture ranges should be f/2.8 - f/22.</span><span>sturdy tripod (metal or plastic doesn't make any difference whatsoever at all in safety - if lightning is so close by, you are in trouble anyway)</span><span>cable release, which can be locked</span><span>Slow-speed film: 100 or 200 ISO
</span></span><span>
3. Why is it important to mentally prepare for photographing lightning?
</span><span>When photographing lightning, it’s important to realize that the conditions you are shooting in are unpredictable and dangerous, and there will always be an element of chance and luck involved. So you should prepare yourself.
</span><span>
4. What time of day should you try to photograph lightning?
</span>Nighttime lightning photography is the easiest <span>type
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5. Why is composition important in lightning photographs?
It boosts or adds drama to your picture. </span>
Answer:
Check the explanation
Explanation:
#include <iostream>
using namespace std;
void hex2dec(string hex_num){
int n = 0;
//Loop through all characters in string
for(int i=0;i<hex_num.size();i++){
//take ith character
char c = hex_num[i];
//Check if c is digit
if(c>='0' && c<='9'){
n = 16*n + (c-48);
}
//Convert c to decimal
else{
n = 16*n + (c-55);
}
}
cout<<hex_num<<" : "<<n<<endl;
}
int main()
{
hex2dec("EF10");
hex2dec("AA");
return 0;
}
The Output can be seen below :
Producers must understand the marginal benefit of making an additional unit which shows the possible gain. Marginal benefit is used in business and economics as a measurement of the change in benefits over the change in quantity. Possible gain is one example of benefit. This measurement provides the relevant measurement of benefits at a specific level of production and consumption.
Answer:
SELECT vendor_number, vendor_name, CONCAT ('street', ' ' , 'city', ' ' , 'state', ' ' , 'zip code') as adress
FROM vendor_directory
ORDER BY vendor_name ASC;
Explanation:
* Suppose <u>vendor_directory</u> is the name of the table from which you extract the data with the SELECT sentence.
Answer:
A. the museum can choose a heuristic approach which will achieve a lossless compression, but they cannot be sure that it is the most efficient compression for each image
C. algorithms for lossless compression exist, so the museum can use those to compress the image
Explanation:
