College Physics

The force is equal to the maximum tension, orF= 3.0×10^6 N. The cross-sectional area isπr^2 = 2.46×10−3m^2. The equation ΔL=^1 Y ...

All quantities exceptΔLare known. Note that the compression value for Young’s modulus for bone must be used here. Thus, (5.33) Δ ...

Figure 5.18Shearing forces are applied perpendicular to the lengthL 0 and parallel to the areaA, producing a deformationΔx. Vert ...

A=πr^2 = 1.77×10−6m^2. (5.42) The value forL 0 is also shown in the figure. Thus, (5.43) F= (80×10^9 N/m^2 )(1.77×10−^6 m^2 ) (5 ...

deformation: drag force: friction: Hooke’s law: kinetic friction: magnitude of kinetic friction: magnitude of static friction: S ...

whereμsis the coefficient of static friction, which depends on both of the materials. • The kinetic friction force fkbetween sys ...

9.The elastic properties of the arteries are essential for blood flow. Explain the importance of this in terms of the characteri ...

Problems & Exercises 5.1 Friction 1.A physics major is cooking breakfast when he notices that the frictional force between h ...

Figure 5.22Part of the climber’s weight is supported by her rope and part by friction between her feet and the rock face. 18.A c ...

34.A 20.0-m tall hollow aluminum flagpole is equivalent in strength to a solid cylinder 4.00 cm in diameter. A strong wind bends ...

6 UNIFORM CIRCULAR MOTION AND GRAVITATION Figure 6.1This Australian Grand Prix Formula 1 race car moves in a circular path as it ...

6 Uniform Circular Motion and Gravitation Many motions, such as the arc of a bird’s flight or Earth’s path around the Sun, are c ...

Δθ=2πr (6.2) r = 2π. This result is the basis for defining the units used to measure rotation angles,Δθto beradians(rad), define ...

FromΔθ=Δrswe see thatΔs=rΔθ. Substituting this into the expression forvgives (6.8) v=rΔθ Δt =rω. We write this relationship in t ...

Take-Home Experiment Tie an object to the end of a string and swing it around in a horizontal circle above your head (swing at y ...

Figure 6.8The directions of the velocity of an object at two different points are shown, and the change in velocityΔvis seen to ...

Example 6.2 How Does the Centripetal Acceleration of a Car Around a Curve Compare with That Due to Gravity? What is the magnitud ...

The term rev/min stands for revolutions per minute. By converting this to radians per second, we obtain the angular velocityω. B ...

(6.25) r=mv 2 Fc . This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature ...

(6.30) μs=v 2 rg. Solution for (b) Substituting the knowns, (6.31) μs= (25.0 m/s)^2 (500 m)(9.80 m/s^2 ) = 0.13. (Because coeffi ...