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(III) A bicyclist can coast down a $7.0^{\circ}$ hill at a steady9.5 $\mathrm{km} / \mathrm{h}$ . If the drag force is proportional to the square ofthe speed $v,$ so that $F \mathrm{D}=-c v^{2},$ calculate $(a)$ the value ofthe constant $c$ and $(b)$ the average force that must be appliedin order to descend the hill at 25 $\mathrm{km} / \mathrm{h}$ . The mass of thecyclist plus bicycle is 80.0 $\mathrm{kg}$ . Ignore other types of friction.

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a. $13.723$b. 566.146$N$

Physics 101 Mechanics

Chapter 5

Using Newton's Laws: Friction, Circular Motion, Drag Forces

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

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University of Winnipeg

McMaster University

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In physics, a rigid body is an object that is not deformed by the stress of external forces. The term "rigid body" is used in the context of classical mechanics, where it refers to a body that has no degrees of freedom and is completely described by its position and the forces applied to it. A rigid body is a special case of a solid body, and is one type of spatial body. The term "rigid body" is also used in the context of continuum mechanics, where it refers to a solid body that is deformed by external forces, but does not change in volume. In continuum mechanics, a rigid body is a continuous body that has no internal degrees of freedom. The term "rigid body" is also used in the context of quantum mechanics, where it refers to a body that cannot be squeezed into a smaller volume without changing its shape.

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In physics, rotational dynamics is the study of the kinematics and kinetics of rotational motion, the motion of rigid bodies, and the about axes of the body. It can be divided into the study of torque and the study of angular velocity.

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A bicyclist can coast down…

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A bicyclist of mass 75 $\m…

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(III) A cyclist intends to…

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In this problem, you have a seven degree in fine a study 9.5 kilometers per hour and a drag force that is proportional to *** C B squared and where else a couple e the value of C and the drag force necessary to travel at a higher speed of 25 kilometers per hour. Now because your movie had a constancy, that means you're not accelerating. So if we draw a free body diagram, you have a component of gravity that into you being Angie poolside data and you have a component of gravity that's parallel to be. So you know that MG side data must be equal to my French imports in order for me to move at a constant velocity. And they told me that my purchase boards is proportional to see now. I already know that it's negative because point opposite my direction of motion so I could just leave it a positive mg site. Then we want to solve ports, So if I have a massive 80 gravity off 9.8 sign of seven degrees and that 9.5 kilometers per hour, that's 9000 500 meters divided by 3600 seconds gives us 2.6 38 repeating meters per second. We can use that as my velocity squared Thio get a value for C of 13 point ese. Now we know my value for see. I use that in part B. Now we are moving at a higher constant velocity. So my forces air still equal to each other. But now if I move at a higher positive velocity, that means unless it had a horse in the direction of motion in order from higher speed. So I've been saying that MG science data plus this force is equal to my frictional force, which he said is C B squared. Now we want to solve for this new five force that mg Cynthia minus C Square and my you velocities 25 kilometers per hour and converting that two meters per second. And then we can actually just plug that hole number in on envoy to use. That's 80 9.8 sign of seven degrees minus that see value that will solve for earlier about 13.8. My new velocity, which waas 2.5000 divided by 300 squared, gives me a force off 560 9.96 Mittens

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