Is 303 inch cartridge GB 53 7 collectible?
The .303 cartridge is the standard caliber for the English made Enfields. the other GB 53 7 stands for where and when it was made. These are military designations. Commercial ammuntion would have the .303 caliber stamped on the bottom of the casing along with either the name or initials of the company manufacturing the bullet. The caliber was used by the English for many years including World War II. Most of the NATO nations now use the American .223 caliber in the M14 - M15 - M16. The cartridge can be a collectible however it's value is low in the cartridge collections of today. It is a nice military collectible. Captain Bob email@example.com
76 inch * 53 inch = 73/12 ft * 53/12 ft = 27.9722... sq ft
There are 2.54 centimetres in one inch, therefore, 53 inches is equal to 53 x 2.54 = 134.62 centimetres.
43 out of 303 or about 14.19%
There are 20.86614173 inches in 53 centimeters. 53 centimeters x 1 inch/2.54 centimeters = 20.86614173 inches 1 inch = 2.54 centimeters
53 inches is 134.62cm (There are 2.54 centimeters per inch).
There are 0.0254 metres in one inch. Therefore, 53 inches is equal to 53 x 0.0254 = 1.3462 metres.
There are 2.54 centimetres in one inch. Therefore, 53 inches is equal to 53 x 2.54 = 134.62 centimetres.
There are 25.4 millimetres in one inch. Therefore, 53 inches is equal to 53 x 25.4 = 1346.2 millimetres.
C = ~166.5 inches.
One inch is equal to 2.54 centimetres. Therefore, rounded to two decimal places, 53 centimetres is equal to 53 / 2.54 = 20.87 centimetres.
Yes. You can use the whole number 1 and then follow it with 53/250. OR, you could use 303/350. Hope this helps.
There are 25.4 millimeters to an inch. That works out to 1346.2 millimeters.
1 inch = 25.4 mm thus 53 mm = 53/25.4 = 2.087 ( to 3dp) inches.
6 Nm = 53 in-lb Generic: Formula: Nm * 8.85074579 = Inch-Pounds (=in-lb)
The formula for this is pi x r2, where r = radius, and pi is 22/7 or whatever approximation you use. Your R is 1/2 of 106 = 53 inch. so the area of your circle is pi x r2 = pi x (53) x 53. The math I'll leave up to you.
A 53 inch tall, 9 year old boy should weigh between 70 pounds and 99 pounds. Obesity for this height would begin at 125 pounds.
Information not in the public domain.
There are 2.54 centimeters in one inch. That works out to 134.62 centimeters.
1 inch = 25.4 mm, 53 mm = 2.087 ins Direct Conversion Formula53 mm* 1 in 25.4 mm = 2.086614173 in
Randy Moss is rumored to have a 51 inch verticle jump from a standing position, and a 53 inch verticle jump from a running start.
1 inch = 2.54 centimeters 53 cm * (1 in./2.54 cm) = 20.87 in. 50 centimeters is equal to approximately 20.9 inches.
About 21 inches (20.866 inches) There are 2.54 centimeters in one inch.
1 foot = 12 inches therefore 53 feet 4 inches = 640 inches 1 inch = 0.0254m (or 2.54cm) therefore 53 feet 4 inches (640 inches) = 16.256m
.750" 2.12 Centimeters = 1 Inch 2.12/4=.53 So .53cm = 1/4" 1.59/.53=3 3*.25=.75
AFAIK, that data has not been published.
1 inch = 2.54 centimeters 53 inches = 134.62 centimeters Algebraic Steps / Dimensional Analysis Formula53 in* 2.54 cm 1 in = 134.62 cm
The biggest you can get is: Cab: 160 inch Sleeper Trailer 53 feet 102 iches 14 feet
It is different with every helicopter. For example, the Blackhawk rotor is 53 ft 8 inch in diameter.
98 chevy s10
70-78 pounds :)
53 inches divided by 39.4, call it 1.345 metres.
It depends on the size of the tires--you'll get fewer 22.5-inch tires in a trailer than you will 13-inch ones--but I've hauled 800 tires in one load.
Mikhail Lomonosov was born on November 19, 1711 and died on April 15, 1765. Mikhail Lomonosov would have been 53 years old at the time of death or 303 years old today.
Find the speed of a vehicle whose wheels are 20 inches in diameter are making 53 rotations per minute.?
each rotation = circumference = (pi) x (diameter) = 20 pi inches (53 rotation/minute) x (20 (pi) inch/rotation) x (60 minute/hour) x ( 1 foot/12 inch) x (1 mile/5,280 feet) = (53 x 20 pi x 60)/(12 x 5,280) = 3.153 miles/hour (rounded)
30 (two rows of 15 if the fork entry is on the 48 inch side and two 48 inch pallets can fit next to each other in the trailer). or 26 (two rows of 13 if the fork entry is on the 40 inch side). The above assumes one layer of pallets (no stacking).
53 and 1/3 paving slabs exactly.
Some of the Penny - Farthing bicycles had a 60-spoke 53-inch front wheel and a 20-spoke 18-inch rear wheel . Different manufacturers used different dimensions . the frount wheel is 1.50m and the small wheel is 50m
The year MUST start on a Sunday. For a leap year, it can start on Saturday or Sunday. In any period of 400 years there are 303 non-leap years, of which 43 begins and ends with a Sunday, and there are 97 leap years, of which 28 begins with a Saturday or a Sunday. So the probability in a non-leap year is 43/303, or 14.2%. And the probability in a leap year is 28/97, or… Read More
Formula for the volume of a sphere = 4/3*pi*radius3 Volume = 4/3*pi*53 => 523.5987756 cubic inches
Around $300-$400 if in Very good condition in original 6,5 X 53 Dutch caliber with all numbers matching. If it's been converted to British .303 calber (Dutch East Indies) the price goes down.
For Class 8 truck pulling a 53' trailer, you can expect the length to typically be between 60 and 80 feet, depending on the wheelbase and hood extension of the power unit. With a Volvo VNL730 (with a 233 inch wheelbase) and 48' flatbed with the fifth wheel slid all the way to the rear, I measure out around 67 feet. With a Kenworth W900L (with a 270 inch wheelbase) and a 53' stepdeck trailer… Read More
For a 53' dry or refrigerated van trailer, you'd typically be looking at an overall height of 13'6 with a 110 inch inside height (for a standard cube van - when you get into single or double drop vans, the inside height changes considerably due to the lower deck height), and overall width of 102" with 98 to 101 inches of usable space inside the cube, and 53 feet of length.
You would need 54 tiles. The area is just over 53 tiles in coverage.
Most CRT-style projection TV's have an alignment / focus procedure accessed by the remote control or through buttons on the set itself.
There is no size limit to the first baseman or catcher mitt in high school baseball. The size limit of a high school outfielder's mitt shall be no larger than 14-inches in height, 8-inch width at palm, and 53/.4-inch webbing.
10% off of 53 = 10% discount applied to 53 = 53 - (10% * 53) = 53 - (0.10 * 53) = 53 - 5.3 = 47.7
53 percent = 53/100 in fraction 53% = 53%/100% = 53/100 in fraction
53% is a little over half. 53% of 100 is 53. 53% of 10 is 5.3. 53% of 200 is 106.
from 1948 through 1952 there were 2 size motors a 61 cu. and a 74 cu . from 53 on i believe they were all 74 cu