In lecture demonstrations, we do
measurements of the drag force on different objects. The objects are placed in a uniform airstream created by a fan. Calculate the Reynolds number and the drag coefficient.
The calculus of velocity-dependent frictional forces
When a body slides across a surface, the frictional force on it is approximately constant and given by
${\mu}_{\text{k}}N.$ Unfortunately, the frictional force on a body moving through a liquid or a gas does not behave so simply. This drag force is generally a complicated function of the body’s velocity. However, for a body moving in a straight line at moderate speeds through a liquid such as water, the frictional force can often be approximated by
${f}_{R}=\text{\u2212}bv,$
where
b is a constant whose value depends on the dimensions and shape of the body and the properties of the liquid, and
v is the velocity of the body. Two situations for which the frictional force can be represented this equation are a motorboat moving through water and a small object falling slowly through a liquid.
Let’s consider the object falling through a liquid. The free-body diagram of this object with the positive direction downward is shown in
[link] . Newton’s second law in the vertical direction gives the differential equation
$mg-bv=m\frac{dv}{dt},$
where we have written the acceleration as
$dv\text{/}dt.$ As
v increases, the frictional force –
bv increases until it matches
mg . At this point, there is no acceleration and the velocity remains constant at the terminal velocity
${v}_{\text{T}}.$ From the previous equation,
$mg-b{v}_{\text{T}}=0,$
so
${v}_{\text{T}}=\frac{mg}{b}.$
We can find the object’s velocity by integrating the differential equation for
v . First, we rearrange terms in this equation to obtain
$\frac{dv}{g-(b\text{/}m)v}=dt.$
Assuming that
$v=0\phantom{\rule{0.2em}{0ex}}\text{at}\phantom{\rule{0.2em}{0ex}}t=0,$ integration of this equation yields
where
$v\text{'}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}t\text{'}$ are dummy variables of integration. With the limits given, we find
Since
$\text{ln}A-\text{ln}B=\text{ln}(A\text{/}B),$ and
$\text{ln}(A\text{/}B)=x\phantom{\rule{0.2em}{0ex}}\text{implies}\phantom{\rule{0.2em}{0ex}}{e}^{x}=A\text{/}B,$ we obtain
A motorboat is moving across a lake at a speed
${v}_{0}$ when its motor suddenly freezes up and stops. The boat then slows down under the frictional force
${f}_{R}=\text{\u2212}bv.$ (a) What are the velocity and position of the boat as functions of time? (b) If the boat slows down from 4.0 to 1.0 m/s in 10 s, how far does it travel before stopping?
Solution
With the motor stopped, the only horizontal force on the boat is
${f}_{R}=\text{\u2212}bv,$ so from Newton’s second law,
$m\frac{dv}{dt}=\text{\u2212}bv,$
which we can write as
$\frac{dv}{v}=-\frac{b}{m}dt.$
Integrating this equation between the time zero when the velocity is
${v}_{0}$ and the time
t when the velocity is
$v$ , we have
As time increases,
${e}^{\text{\u2212}bt\text{/}m}\to 0,$ and the position of the boat approaches a limiting value
${x}_{\text{max}}=\frac{m{v}_{0}}{b}.$
Although this tells us that the boat takes an infinite amount of time to reach
${x}_{\text{max}},$ the boat effectively stops after a reasonable time. For example, at
$t=10m\text{/}b,$ we have
Therefore, the boat’s velocity and position have essentially reached their final values.
With
${v}_{0}=4.0\phantom{\rule{0.2em}{0ex}}\text{m/s}$ and
$v=1.0\phantom{\rule{0.2em}{0ex}}\text{m/s,}$ we have
$1.0\phantom{\rule{0.2em}{0ex}}\text{m/s}=(4.0\phantom{\rule{0.2em}{0ex}}\text{m/s}){e}^{\text{\u2212}(b\text{/}m)(10\phantom{\rule{0.2em}{0ex}}\text{s})},$ so
branch of science dt deals with the study of physical properties of matter and it's particulate nature
Josiah
Good
Daniel
actually
Nathz
Y acctually do u hav ur way of defining it? just bring ur iwn idear
Daniel
well, it deals with the weight of substances and reaction behind them as well as the behavior
Josiah
buh hope Esther, we've answered ur question
Josiah
what's ohms law
CHIJIOKE
ohms law states that, the current flowing through an electric circuit is directly proportional to the potential difference, provided temperature and pressure are kept constant
Josiah
what is sound
James
ohms law states that the resistance of a material is directly proportional to the potential difference between two points on that material, if temperature and other physical conditions become constant
vector quantity is any quantity that has both magnitude in terms of number (units) and direction in terms of viewing the quantity from an origin using angles (degree) or (NEWS) method
LEWIS
vector quantity is physical quantity has magnitude and direction
vector is a quantity that is use in measuring size of physical properties and their direction
Bitrus
what difference and similarities between work,force,energy and power?
enery is the ability to do work. work is job done, force is a pull or push. power has to do with potential. they belong to different categories which include heat energy, electricity.
Andrew
force refers to a push or pull... energy refers to work done while power is work done per unit time
Shane
mathematically express angular velocity and angular acceleration
it depends on the direction. an angular velocity will be linear and angular acceleration will be an angle of elevation.
Andrew
The sonic range finder discussed in the preceding question often needs to be calibrated. During the calibration, the software asks for the room temperature. Why do you suppose the room temperature is required?
Suppose a bat uses sound echoes to locate its insect prey, 3.00 m away. (See [link] .) (a) Calculate the echo times for temperatures of 5.00°C5.00°C and 35.0°C.35.0°C. (b) What percent uncertainty does this cause for the bat in locating the insect? (c) Discuss the significance of this uncertainty an
Shaina
give a reason why musicians commonly bring their wind instruments to room temperature before playing them.
Shaina
The ear canal resonates like a tube closed at one end. (See [link]Figure 17_03_HumEar[/link].) If ear canals range in length from 1.80 to 2.60 cm in an average population, what is the range of fundamental resonant frequencies? Take air temperature to be 37.0°C,37.0°C, which is the same as body tempe
Shaina
By what fraction will the frequencies produced by a wind instrument change when air temperature goes from 10.0°C10.0°C to 30.0°C30.0°C ? That is, find the ratio of the frequencies at those temperatures.