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(II) The terminal velocity of a $3 \times 10^{-5} \mathrm{kg}$ raindrop is about9 $\mathrm{m} / \mathrm{s} .$ Assuming a drag force $F_{\mathrm{D}}=-b v,$ determine $(a)$ thevalue of the constant $b$ and $(b)$ the time required for such adrop, starting from rest, to reach 63$\%$ of terminal velocity.

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a. $3 \times 10^{-5} k g / s$b. 1$s$

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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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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so the terminal velocity has given an equation 5.9. And from that equation we see that the terminal velocity is mg Overby. And from here we can solve for the constantly, which is nothing but mg over CT. Now let's use the given numbers for M. We had three times into the part minus five. Cagey and ah G is 9.8 meet over seconds. Word also HVT is nine meters per second. So combining everything together receive that determines the constant is 3.27 times tend to the bar minus five kg per second s So that's part E for part B. We need to figure out the time constant um, all the time required for such a drop, starting from rest to 63% of the terminal velocity. And to do so, we can use example 5.17 where the time required for the velocity to read 63% of the terminal velocity is basically the time constant. How on and that's am overby. So I'd highly recommend you check Example five points. 17 Where, um, it's calculated how it's given how to calculate the time constant or the time required for the velocity to read 63% off the terminal velocities. In other words, that's towels. And, uh, if we now use the even numbers, we see that it's three times 10 to the powers minus five kg, divided by B, which is 3.27 times 10 power buying his five kg per second. And that gives us 0.917 seconds, which is roughly one thank you.

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