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3 years ago

The term meniscus refers to

1 answer:

3 0

PHYSICSthe curved upper surface of a liquid in a tube.OPTICSa lens that is convex on one side and concave on the other.ANATOMYa thin fibrous cartilage between the surfaces of some joints, e.g., the knee.

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2 years ago

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Given A=125.0=0.4 and 25.0=0.1 calculate A-B

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2 years ago

Just a thought

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THE FOURTEEN WEEKS' COURSES

NATURAL SCIENCE,

J. DORMAN STEELE, A.M., PHI.D.

Fourteen Weeks iq Natural Philosophy,

Fourteen Weeks iq Ctlenqistry,

Fourteen Weeks iq Descriptive Astroqonqy,

Fourteel Weeks iq Popular Geology,

Fourteeq Weeks iQ2 Human P1ysiology,

Fourteen Weeks iq Zoology,

Fourteeq Weeks iq Botany,

A Key, containing Answers to the Questions

and Problems in Steele's I4 Weeks' Courses,

4 1ISTORIC4L SERIES,

ON THE PLAN OF STEELE'S 14 WEEKS IN THE SCIENCES.

A Brief History of the Urlited States,

A Brief History of France,

The same publishers also offer the following standard scientific

vworks, being more extended or difficult treatises than those of

Prof. Steele, though still of Academic grade.

Peck's Ganot's Natural Philosophy,

Porter's Principles of Chemistry,

Jarvis' Physiology and Laws of Healtl,

Wood's Botanist and Florist,

Clanlbers' Elenments of Zoology,

lcIqtyre's Astroqomy and tle Globes,

Page's. Elen~ents of Geology,

Entered according to Act of Congress, in the year 1869, by

A. S. BARNES & CO.,

In the Clerk's Office of the District Court of the United States

for the Southern District of New York.

sTERLE'S KEY.

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2 years ago

A 0.80-μm-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet hangs motionless if t

Arisa [49]

A) $2.4\cdot 10^{-16}kg$

The radius of the oil droplet is half of its diameter:

$r=\frac{d}{2}=\frac{0.80 \mu m}{2}=0.40 \mu m = 0.4\cdot 10^{-6}m$

Assuming the droplet is spherical, its volume is given by

$V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.4\cdot 10^{-6} m)^3=2.68\cdot 10^{-19} m^3$

The density of the droplet is

$\rho=885 kg/m^3$

Therefore, the mass of the droplet is equal to the product between volume and density:

$m=\rho V=(885 kg/m^3)(2.68\cdot 10^{-19} m^3)=2.4\cdot 10^{-16}kg$

B) $1.5\cdot 10^{-18}C$

The potential difference across the electrodes is

$V=17.8 V$

and the distance between the plates is

$d=11 mm=0.011 m$

So the electric field between the electrodes is

$E=\frac{V}{d}=\frac{17.8 V}{0.011 m}=1618.2 V/m$

The droplet hangs motionless between the electrodes if the electric force on it is equal to the weight of the droplet:

$qE=mg$

So, from this equation, we can find the charge of the droplet:

$q=\frac{mg}{E}=\frac{(2.4\cdot 10^{-16}kg)(9.81 m/s^2)}{1618.2 V/m}=1.5\cdot 10^{-18}C$

C) Surplus of 9 electrons

The droplet is hanging near the upper electrode, which is positive: since unlike charges attract each other, the droplet must be negatively charged. So the real charge on the droplet is

$q=-1.5\cdot 10^{-18}C$