Circle, diameter radius

RADIUS of a circle is the line segment whose one endpoint is at the center of the circle and whose other endpoint is on the circle. The radius is one-half the length of the diameter r= d. 

Calculate the area (in m2) of a circle (circle 1) that has a diameter of 10.0 cm. Calculate the area (in m2) of a circle (circle 2) that has a diameter of 5.00 cm. Now, calculate the ratio (area of circle l)/(area of circle 2), and compare it to the ratio of (radius of circle l)/(radius of circle 2). What effect does cutting the diameter by a factor of 2 have on the area ... 

Area of a circle K radius squared or diameter squared x 0.7854... 

Fig. 7. Scattering curves from models of uniform scattering density, (a) A sphere of radius 6.5 nm and Rq 5.0 nm as used in Fig. 4. (b) A hollow sphere with an outer radius of 6.5 nm as in (a) and a ratio of inner/outer radii of 0.5. (c) A straight, cylindrical rod of length 59 nm and diameter 3.4 nm. (d) The cylindrical rod as in (c) is now bent into a circle of radius 9.4 nm.

The Mohr circle representation (Fig. 9.6c) is a graphical method of relating stress components in different sets of axes. When the axes in the material rotate by an angle B, the diameter of the circle rotates by an angle 2 B. If the material yields, the circle has radius k, the constant in the Tresca yield criterion. The axes of the Mohr diagram are the tensile and shear stress components. Thus, in the left-hand circle, representing the stresses at A in Fig. 9.6b, the ends of the horizontal diameter are the principal stresses. The principal axes are parallel and perpendicular to the notch-free surface. There is a tensile principal stress Ik parallel to the surface, and a zero stress perpendicular to the surface. The points at the ends of the vertical diameter represent the stress components in the a)3 axes, rotated by 45° from the principal axes. In the a/3 axes, the shear stresses have a maximum value k, and there are equal biaxial tensile stresses of magnitude = k (the coordinate of the centre of the circle). 

Mark off the length of the circle s radius from the point the two perpendicular lines intersect (the radius will be one-half the diameter of the eventual circle). For this example, that will be one-half inch, make the diameter of the circle 1 inch (see Figure 4-20, photo 4). 

If a cone is unfolded while pivoting about the apex O, the development is a segment of a circle of radius Oa whose arc ab is equal in length to the circumference of the base. Fig. 4.11. To find the length of arc ab, the base diameter is equally divided into 12 parts. The 12 small arcs 1-2, 2-3, etc. are transferred to the arc with point 12 giving the position of point b. A part cone (frustum) is developed in exactly the same way, with the arc representing the small diameter having a radius Oc, Fig. 4.12. 

One way of visualizing SHM is to im ine a point rotating around a circle of radius rat a constant angular velocity to. If the distance from the centre of the circle to the projection of this point on a vertical diameter is y at time t, this projection of the point will move about the centre of the circle with simple harmonic motion. A graph of y gainst t will be a sine wave, whose equation is y = isintot (see diagram). See also pendulum. 

Grid Hexagon Leg (pm) Square Side (pm) Circle Diameter (pm) Peanut Radius (pm)... 

Consider Figure 4.1. A particle, of mass i. rotates at a tangential velocity. >V. and angular velocity, u . in a Circle of radius, r. After a time, t. the particle has moved to a point on the circle radius, r, which subtends an angle. 7. where 7 — ujl, from its position at time i-O. the extreme right of the horizontal diameter of the circle. 

To find expressions for X and z, we need a slightly more elaborate version of the kinetic model of gases. The basic kinetic model supposes that the molecules are effectively pointlike however, to obtain collisions, we need to assume that two points score a hit whenever they come within a certain range d of each other, where d can be thought of as the diameter of the molecules (Fig. 7.20). The collision cross-section, a (sigma), the target area presented by one molecule to another, is therefore the area of a circle of radius d, so O = nd. When this quantity is built into the kinetic model, we find that... 

Fig. 7.20 To calculate features of a perfect gas that are related to collisions, a point is regarded as being surrounded by a sphere of diameter d. A molecule will hit another molecule if the center of the former lies within a circle of radius d.

The first solution for folding the optical path is the Herriot delay line, with two concave mirrors having a radius of curvature such that the injected light exits from the entrance hole after N round-trips, which means after an optical path 2NL. The number of reflections is limited by the reflection losses, thus the mirrors should have high optical quality over the whole surface. During the N trips the laser spots move over a circle with radius Ncofj, where laser beam width in typical working conditions, the mirrors should be 1 m in diameter, a number with a great impact on the scale and the cost of the vacuum system. 

A chord of a circle (or sphere) is a line segment whose end points lie on the circle (or sphere). A line which intersects the circle (or sphere) in two points is a secant of the circle (or sphere). A diameter of a circle (or sphere) is a chord containing the center and a radius is a line segment from the center to a point on the circle (or sphere). 

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