All categories

2 years ago

Just a thought

Physics

1 answer:

Olin [163]2 years ago

6 0

Answer:

Explanation:

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.

You might be interested in

What is a photovoltaic cell

noname [10]

Its a solar cell. Photo voltaic Cell is also known as a solar cell.

4 0

3 years ago

The lowest possible temperature in outer space is 3.13 K. What is the rms speed of hydrogen molecules at this temperature? (The

Umnica [9.8K]

Answer:

$v_{rms} =196.59 m/s$

Given:

Temperature, T = 3.13 K

molar mass of molecular hydrogen, m = 2.02 g/mol = $2.02\times 10^{-3}kg/mol$

Solution:

To calculate the root mean squarer or rms speed of hydrogen molecule, we use the given formula:

$v_{rms} = \sqrt{\frac{3TR}{m}}$

where

R = rydberg's constant = 8.314 J/mol-K

Putting the values in the above formula:

$v_{rms} = \sqrt{\frac{3\times 3.13\times 8.314}{2.02\times 10^{-3}}}$

$v_{rms} =196.59 m/s$

5 0

3 years ago

A 1-mCi source of^60 Co is placed in the center of a cylindrical water-filled tank with an inside diameter of 20 cm and depth of

LenKa [72]

To solve this problem we need the concepts of Energy fluency and Intensity from chemical elements.

The energy fluency is given by the equation

$\Psi=4RcE\pi$

Where

$\Psi =$ The energy fluency

c = Activity of the source

r = distance

E = electric field

In the other hand we have the equation for current in materials, which is given by

$I= I_0 e^{-\mu_{h20}X_{h2o}} e^{-\mu_{Fe}X_{Fe}}$

Then replacing our values we have that

$I = 1*10^{-3} * 3.3*10^{10} * e ^{-0.06*1.1} e^{-0.058*7.861}$

$I = 1.3*10^7 Bq$

We can conclude in this part that 1.3*10^7Bq is the activity coming out of the cylinder.

Now the energy fluency would be,

$\Psi = \frac{cE}{4\pir^2}$

$\Psi = \frac{1.3*10^7*2*1.25}{4\pi*11^2}$

$\Psi = 2.14*10^4 MeV/cm^2.s$

The uncollided flux density at the outer surface of the tank nearest the source is $\Psi = 2.14*10^4 MeV/cm^2.s$

6 0

2 years ago

8. List three temperature scales. Then write the boiling and freezing

Tomtit [17]

Explanation:

For most temperature scales, the boiling point of water and the freezing point is used to calibrate it.

Three known temperature scales;

Kelvin scale
Celcius scale
Fahrenheit scale

Kelvin scale Celcius scale Fahrenheit scale

Freezing point 273K 0°C 32°F

Melting point 373K 100°C 212°F

5 0

2 years ago

A rectangular plate, whose streamwise dimension (or chord c) is 0.2 m and whose width (or span b) is 1.8 m, is mounted in a wind

34kurt

Answer:

Answer for the question is given in the attachment.

Explanation:

Download pdf

4 0

3 years ago