The length of an arc, with an angle in degrees, is equal to (pi x r x θ)/180.In this case, it is (pi x 120 x 10)/180, which is (20pi)/3 or about 20.944.This answer is not right for A+
150 degrees
The triangle ABC is an equallateral triangle since angle ABC is one sixth of 360 degress of the circle and the angles BAC and BCA are equal of the remaining 180-60=120 degrees. With radius BC (or BA) being 6; the areaof the circle is pi (r)squared; 36 piArea of the circle is 36piMalcolm Lowe
14 units.
Using only a compass you cannot do it - you will need a straight edge as well.Draw a straight line AB.From A, using the length AB draw an arc above AB.From B, using the length AB draw an arc to intersect the first arc at C.Join CA.Then angle CAB is 60%From A, draw an arc to cut the two arm of the angle, AB and AC at X and Y respectively.From X, draw an arc between the two arms of angle CAB.Using the same length, repeat from Y so as to cut the previous arc at D.Join DA.DA bisects angle CAB and so angle DAB = 30 degrees.Now bisect angle DAB to give angle EAB = 15 degrees.At A, draw a perpendicular (facing downwards). If F is any point on the perpendicular, the angle EAF = 15 + 90 = 105 degrees.
The length of an arc, with an angle in degrees, is equal to (pi x r x θ)/180.In this case, it is (pi x 120 x 10)/180, which is (20pi)/3 or about 20.944.This answer is not right for A+
It depends on where arc AC is.
The 6013 Rod can be used on either ac or dc arc welders, at around 90-120 amps depending on your settingsTucson Direct
The sides are all 10. The vertex is a right angle. This makes the diagonal formed by ac the hypotenuse of a right triangle with sides 10 and 10. 10^2+10^2=ac^2 100+100=ac^2 200=ac^2 10* sqr 2=ac approximately 14.1421....=ac
ac = b ac = 120 So, 1,2,3,4,5,6,8,10,12,15,20,24,30,40,60, and 120 are the divisors or factors of 120 Also we can express 120 as: 120 = 2^3 x 3 x 5
yes, but you must derate the voltage as DC arc across blown fuse is not self quenching as is AC arc.
10 amps at 120 volts is 1200 watts or 1.2 Kw, so in 1 hour it will use 1.2 Kwh
Draw a horizontal line AB equal to one of the side lengths. From A draw an arc of a circle of radius one of the remaining lengths. From B draw an arc of a circle of radius the third length. Where the arcs intersect is point C. Join AC and BC. Voila!
Stick welding is also more difficult to learn and use, particularly the ability to strike and maintain an arc. Arc welders are available in AC, DC or AC/DC, with AC being the most economical. It's used for welding thicker metals of 1/16 inch or greater.
A circuit breaker is easier to design for ac than dc because alternating current (ac) naturally goes to zero 100 or 120 times per second and this helps to extinguish any arc. Therefore an ac circuit breaker would not be suitable for dc assuming the same voltage and current ratings.
The acceptable range of ac voltage at an outlet is from 115 to 120 volts. The utility company has a mandate to keep the voltage within a 10% fluctuation range. The catch is 10% from what set point voltage.
If it's a right angle triangle then side ac is 10 units in length.