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valentinak56 [21]
1 year ago
12

The great astronomer of ancient times who summarized and improved a system of circles upon circles to explain the complicated mo

tions of the planets (and published the system in a book now called The Almagest) is:
Physics
1 answer:
Ainat [17]1 year ago
6 0

The great astronomer of ancient times who summarized and improved...in a book now called The Almagest) is Ptolemy This is further explained below.

<h3>Who is Ptolemy?</h3>

Generally, Claudius Ptolemy was a Greek mathematician, astronomer, and geographer who lived in the second century CE and is best known for proposing the geocentric model of the cosmos, which was used to explain planetary and stellar movements for the next thousand years.

In conclusion, Ptolemy, the ancient world's preeminent astronomer, compiled and refined a system of circles inside circles to describe the complexities of planetary motion, publishing his work in what is now known as The Almagest.

Read more about Ptolemy

brainly.com/question/15075606

#SPJ1

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The membrane that surrounds a certain type of living cell has a surface area of 5.1 x 10-9 m2 and a thickness of 1.4 x 10-8 m.
ziro4ka [17]

a) The charge on the outer surface is 1.2\cdot 10^{-12} C

b) The number of ions is 7.5\cdot 10^6

Explanation:

a)

The membrane behaves as a parallel plate capacitor, whose capacity is given by the equation

C=\frac{k\epsilon_0 A}{d}

where

k = 4.3 is the dielectric constant

\epsilon_0 =8.85\cdot 10^{-12} F/m is the vacuum permittivity

A=5.1\cdot 10^{-9} m^2 is the surface area

d=1.4\cdot 10^{-8} m is the distance between the two plates

Substituting,

C=\frac{(4.3)(8.85\cdot 10^{-12})(5.1\cdot 10^{-9})}{1.4\cdot 10^{-8}}=1.4\cdot 10^{-11} F

The capacity of the membrane is related to the potential difference between the two surfaces by

C=\frac{Q}{\Delta V}

where here we have

Q = excess charge on one surface

\Delta V = 85.5 mV = 0.0855 V is the potential difference between the two surfaces

Solving for Q, we find

Q=C\Delta V=(1.4\cdot 10^{-11})(0.0855)=1.2\cdot 10^{-12} C

b)

We said that the net charge on the outer surface is

Q=1.2\cdot 10^{-12} C

The charge of one K+ ions is equal to the electron charge

+e=1.6\cdot 10^{-19} C

Therefore, the number of ions on the outer surface can be found by dividing the total charge by the charge of a single ion:

N=\frac{Q}{e}=\frac{1.2\cdot 10^{-12}}{1.6\cdot 10^{-19}}=7.5\cdot 10^6

Learn more about capacity:

brainly.com/question/10427437

brainly.com/question/8892837

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#LearnwithBrainly

6 0
3 years ago
If a gas has a gage pressure of 156 kPa, it is absolute pressure is approximately
Art [367]
In the given question, one important information for getting to the actual solution is not given and that is the atmospheric pressure. To find the approximate absolute pressure, it is needed to add the value of atmospheric pressure with the gage pressure.
Atmospheric pressure = 100 kPa
Then
Absolute pressure = 156 + 100 kPa
                             = 256 KPa.
5 0
3 years ago
How many electrons have been removed from a positively charged electroscope if it has a net charge of 6x10-11?
oksano4ka [1.4K]

Answer:

is it the ans you are looking for

5 0
3 years ago
A weight of 30.0 N is suspended from a spring that has a force constant of 220 N/m. The system is undamped and is subjected to a
Nimfa-mama [501]

Answer:

F_0 = 393 N

Explanation:

As we know that amplitude of forced oscillation is given as

A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}

here we know that natural frequency of the oscillation is given as

\omega_0 = \sqrt{\frac{k}{m}}

here mass of the object is given as

m = \frac{W}{g}

\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}

\omega_0 = 8.48 rad/s

angular frequency of applied force is given as

\omega = 2\pi f

\omega = 2\pi(10.5) = 65.97 rad/s

now we have

0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}

F_0 = 393 N

6 0
3 years ago
Three forces act on an object. Two of the forces are at an angle of 100◦to each other and have magnitude 25N and 12N. The third
seraphim [82]

Answer:

F₄ = 29.819 N

Explanation:

Given

F₁ = (- 25*Cos 50° i + 25*Sin 50° j + 0 k) N

F₂ = (12*Cos 50° i + 12*Sin 50° j + 0 k) N

F₃ = (0 i + 0 j + 4 k) N

Then we have

F₁ + F₂ + F₃ + F₄ = 0

⇒   F₄ = - (F₁ + F₂ + F₃)

⇒   F₄ = - ((- 25*Cos 50° i + 25*Sin 50° j) N + (12*Cos 50° i + 12*Sin 50° j) N + (4 k) N) = (13*Cos 50° i - 37*Sin 50° j - 4 k) N

The magnitude of the force will be

F₄ = √((13*Cos 50°)² + (- 37*Sin 50°)² + (- 4)²) N = 29.819 N

6 0
3 years ago
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