Its calibre is .45 however I presume you want measurements its barrel starts at roughly 3 cm or an 1 inch and a quarter and ends at 1 and a half cm or half an inch...
I'm not taking it apart to find out however its barrel length around 28 cm or 11 and a quarter inches even more taking into account the flash guard 6 and a half cm or 2 and three quarter inches. If you require any more measurements just ask!
length*width = area 24.7*width = 550.81 Divide both sides by 24.7 to find the width: width = 22.3 meters
Let the width of rectangle is w cm (Length is 12 cm more) so length is w + 12 Perimeter = 16 times of width 2 (length + width) = 16 * width Rest of the homework, do yourself.
Length is 12ft, width is 4 ft. 2 x (length + width) = perimeter = 32 ft ⇒ length = 16ft - width [1] length x width = area = 48 sq ft Using [1] above: ⇒ (16ft - width) x width = 48 sq ft multiplying out and rearranging: ⇒ width2 - 16 x width + 48 = 0 factorizing: ⇒ (width - 4)x(width - 12) = 0 Thus width = 4 or 12, so using [1] above: ⇒ width = 4ft, length = 16 - 4 = 12ft or width = 12ft, length = 16 - 12 = 4ft Convention is that the length is larger than the width, thus rectangle is 12ft long by 4 ft wide.
Important to note are these formulae: Perimeter_of_rectangle = 2 x (length + width) Area_of_rectangle = length x width So if the perimeter and area are known, then: 2 x (length + width) = perimeter => length + width = perimeter / 2 => length = perimeter / 2 - width length x width = area => (perimeter / 2 - width) x width = area (substituting for length given above) => perimeter / 2 x width - width2 = area => width2 - perimeter / 2 x width + area = 0 which is a quadratic and can be solved either by factorization or by using the formula: width = (perimeter / 2 +/- sqrt(perimeter2 / 4 - 4 x area)) / 2 = (perimeter +/- sqrt(perimeter2 - 16 x area)) / 4 This will provide two values for the width. However, each of these values is the length for the other, so the larger value is the length and the smaller value is the width. Sometimes only 1 value will be found for the width above. In this case, the rectangle is actually a square which means that the length and width are both the same. Examples: 1. perimeter = 6, area = 2 width2 - perimeter / 2 x width + area = 0 => width2 - 6 / 2 x width + 2 = 0 => width2 - 3 x width + 2 = 0 => (width - 2) x (width - 1) = 0 => width = 2 or 1. So the length is 2 and the width is 1. 2. perimeter = 12, area = 9 width2 - perimeter / 2 x width + area = 0 => width2 - 12 / 2 x width + 9 = 0 => width2 - 6 x width + 9 = 0 => (width - 3)2 = 0 => width = 3 So the rectangle is a square with both length and width of 3.
perimeter 24 so length + width half of that ie 12. Length 3 times width must be 9 and width 3.
Of course they CAN
Costco.
the thompson m1 hast a smaller magazine 20 bullets the m1a1 has 30 dont know rest
M1A1 Thompson, Colt 1911, M1A1 Grand, and M1A1 Carbines (For Foot Solders)
.45 caliber ACP
It all depends of the company making it. A Tokyo Marui M1A1 Thompson would be better than a cheap Wal-Mart M4. But a Classic Army M4 would be better than a CYMA M1A1 Thompson.
M1A1 Thompson's were fed from stick magazines that came in either 20 or 30 round capacity. The most common being the 20 round magazine.
No. It is chambered for the .45 ACP cartridge.
personally i wouldrecommend a m4 as is cheap easy and cheap to fix but a thompson would be fine
4 Types: M1921 M1927/8 M1 M1A1
Walmart is the only place that carries the really cheap plastic guns
The present day semi auto version is marked TM1, and is a copy of the M1A1 submachine gun. During the war, the 1928, M1, and M1A1 were used.