Diameter = 6.6cm, so radius = 3.3cm
Circumference = 2 * pi * radius
Circle. = 20.735 cm
Now we multiply by the length.....
20.735 * 8.8 = 182.468 cm^2
Now we have the 1cm overlap, which technically is a rectangle of dimensions 1cm * 8.8cm i.e. 1 * 8.8cm = 8.8 cm^2
We just add this on.....
182.468 + 8.8 = 191.268 cm^2 or 191.27 cm^2 (to 2dp)
A cylinder with a diameter of 1 foot and 3 feet high has a surface area of 11 square feet.
The volume of a cylinder 2 feet in diameter and 5 feet high is: 15.7 cubic feet.
The volume of a cone 4 feet high with a base diameter of 13 feet is: 177 cubic feet.
The total surface of the base is 2206 square feet (approx). The area of the top, if there is any, will be the same. The area of the side walls is 1582 square feet (approx).
Dimensions are 970mm high by 580mm diameter
A cylinder with a diameter of 1 foot and 3 feet high has a surface area of 11 square feet.
diameter X pi = circumference circumference X height = area
Strange p-gram to have a diameter! Area is 30 x 20 ie 600 cm2
Area = pi*502 = 7853.981634 or about 7854 square feet.
area of object = (1/3) pi * radius^2 = (1/3) (pi) * (0.6)^2 = 0.377 Find the diameter of this object (assuming it's a circle), and that's the answer: diameter = radius * 2 radius = square root (area / pi) diameter = 2 * square root (area / pi) diameter = 2 * (0.335) = 0.67
The diameter of the field of view decreases when changing from low to high power magnification. This is because higher magnification zooms in closer on the specimen, limiting the area of the specimen that can be seen at one time.
When you switch from low to high power on a microscope, the diameter of the field of view decreases because the high-power objective has a higher magnification, which zooms in on a smaller area. This allows for more detailed observation of the specimen at the expense of a smaller field of view.
The volume of a cylinder that is 30mm in diameter and 700mm high is 494,800.84mm3
No, the wire with a diameter of 0.01 mm will have higher resistance compared to a wire with a diameter of 0.1 mm. Resistance of a wire is inversely proportional to its cross-sectional area, so a thinner wire will have higher resistance.
The diameter is 8 feet, circumference 25.12, area 50.24
Low resistance.AnswerSince resistance is inversely proportional to the cross-sectional area of a conductor, increasing the diameter ('thickness') of a conductor will reduce its resistance.For example, doubling the diameter of a circular-section conductor will quadruple its cross-sectional area, and reduce its resistance by one quarter.
The volume of cylinder 1' high x 1' diameter is: 0.785 cubic feet.