In classical geometry, a radius of a circle or sphere is any of the line segments from its center to its perimeter, and in more modern usage, it is also their length. The name comes from the Latin radius, meaning ray but also the spoke of a chariot wheel. The plural of radius can be either radii (from the Latin plural) or the conventional English plural radiuses. The typical abbreviation and mathematical variable name for radius is r. By extension, the diameter d is defined as twice the radius:
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{\displaystyle d\doteq 2r\quad \Rightarrow \quad r={\frac {d}{2}}.}
If an object does not have a center, the term may refer to its circumradius, the radius of its circumscribed circle or circumscribed sphere. In either case, the radius may be more than half the diameter, which is usually defined as the maximum distance between any two points of the figure. The inradius of a geometric figure is usually the radius of the largest circle or sphere contained in it. The inner radius of a ring, tube or other hollow object is the radius of its cavity.
For regular polygons, the radius is the same as its circumradius. The inradius of a regular polygon is also called apothem. In graph theory, the radius of a graph is the minimum over all vertices u of the maximum distance from u to any other vertex of the graph.The radius of the circle with perimeter (circumference) C is
TL;DR Summary: For a transiting exoplanet, we find it takes 4.3 days from the start of the transit for the host star to reach a minimum brightness, which lasts for 10 days. Show that the radius of the exoplanet is about 1/2 of Earth’s radius if its orbital radial velocity is 17 m/s.
I am...
"Matter tells spacetime how to curve, and curved spacetime tells matter how to move"
Taking the above into account, the space on the Earth's surface is slightly curved. What is the radius of curvature of space on the Earth's surface due to its mass?
I know there are videos etc explaining why but I thought I would try to find a way to understand this myself.
Imagine a version of πr^2, but instead of being for the area of a circle, it’s for the area of a square.
Call it sqi ar ^2
sqi = the ratio of the ‘diameter’ of the square to the...
Hello, I need some help regarding how to calculate the needed radius to achieve 1 x 10/6 v/m on the surface of a spherical inner grid, inside a conductive vacuum chamber. I have used various equations, however I don’t know if they are very good and would like to know other methods. Thanks!
Four points lie on the surface of a sphere. Given the six distances between the points, calculate the radius of the sphere.
This is (allegedly) an advanced high school level problem. However, it is a remembered problem, so it is possibly misremembered (i.e. there might have been some “bice...
In order to change the radius, additional energy is required, the total energy of mass m on a circular orbit is given by:
$$E_{total} = - \frac {GMm} {2R},$$
The change in energy between orbits ##R## and ##R_{0}## is:
$$\Delta E_{total} = \frac {GMm} {2} \cdot \left( \frac 1 R_{0} - \frac 1 R...
Although 'undefined' it may be reasonable to assume that the hypersphere radius and circumference will respond to the normal relationship of a sphere i.e. 2 times Pi times r. Reasonable I suggest because the 3d surface of a hypersphere is a sphere.
To get to my point, time and distance can be...
A. Correct answer is radius = 1770m, acceleration = 2.73*10^-3m/s.
B. I don't know how to approach this problem. I don't know if I should start with forces, energy, or basic kinematics.
I tried the square root of ((2)(6.67*10^-11)(3.90E+30))/(5.70E+7)
I got 1.55*10^-5 and that is wrong. Maybe I am using the wrong equation but this is the one of professor gave me and I don't know what I am doing wrong :-(
Hi,
I am working with leaf springs and studying the derivation of the formula for the deflection of such a structure. The derivation is shown here:
My only doubt is how to obtain the following formula: $$\delta=\frac{L^{2}}{8R}$$ where: ##\delta## - deflection, ##L## - length of the beam...
So I thought I knew how to do this problem but I've run into some issues that make the algebra feel impossible and I am beginning to feel like I'm taking the wrong approach, I ended up rewriting it in a doc because I was concerned maybe my handwriting was the cause of my error so the work is...
Hello,
This question, which I found in various electricitiy and magnetism books (e.g. Introduction to electrodynamics grif.).
There are many variations of this question, I am mainly interested in the following setup of it:
-Suppose there is a charged disk of radius R lying in the xy-plane, and...
https://www.msn.com/en-us/news/technology/the-astonishing-scientific-theory-that-says-the-universe-might-be-inside-a-black-hole/ar-AA17lxtF?ocid=msedgdhp&pc=U531&cvid=272cb184de9c48fbbd3b321120e37dac
Michio Kaku has often joked, "If you want to know what it looks like inside of a black hole...
The water level in a spherical bowl has a diameter of 30 cm. If the horizontal diameter of the bowl is 10 cm below the water level, calculate the radius of the bowl and the depth of the water in the bowl.
I managed to draw a diagram below:
In my drawing, I am seeing the sphere ABCD as the...
I'm reading the Feynman lectures chapter on "Curved Space", section 6-2. Say we're trying to figure out a way to measure average curvature on Earth. We know that 3d space is curved if Euclidean geometry rules don't work - e.g. the ratio of circumference and radius of a circle isn't ##2\pi##, or...
The answer key states that the new tangential speed is half the original speed. However, this isn't correct right? It should double.
My proof:
##F_c = \frac {mv^2} R##
##F_c = F_t##
##\frac {mv^2} {\frac R 4} = \frac {m(2v)^2} R## If centripetal force were to stay constant.
As such, tangential...
How did the first scientists figure out the orbital radius of a moon of Jupiter? How can observations lead to a calculation of the orbital radius of a moon of another planet?
My approach is as follows;
##\dfrac{dr}{dt}=\dfrac{dr}{da}⋅\dfrac{da}{dt}##
##\dfrac{dr}{da}=\dfrac{1}{2πr}##
##\dfrac{dr}{dt}=\dfrac{1}{2πr}⋅15##
##\dfrac{dr}{dt}=\dfrac{7.5}{πr}##
and noting that ##r= \sqrt{\dfrac {50}{π}}##
##\dfrac{dr}{dt}=\dfrac{7.5}{12.533}=0.598## cm/s
*kindly note...
Probably, to satisfy the significant digits rule for division, we should consider ##r = 5.0 \div 2.0##. But I'm unable to come up with a reason why significant digits rule should not apply to ##r= d \div 2##. Also, if we apply significant digits rule to this calculation then we loose accuracy...
I understood that the event horizon is a null surface and not a place in space, what is the relationship between it and the Schwarzschild radius? Also, what does the Schwarzschild radius physically represent for example for an object such as a star?
Soo. I think this problem is too direct and easy so I think I got it in wrong way: p=h/r and then plug in the K and V and then we get E=E(r) and get derivative and we have minimum? What do you think? is there sth I am missing?
I'm attempting to write some code using the Ruby programming language that will give me the radius of an arc but the only pieces of information I have to work with are the arc length (L), the chord length (C), and the circular segment angle in degrees (A):
I'm hoping someone can show me how to...
$$I = \int{r^2dm}$$
$$dm = \sigma dV$$
$$dV = 4\pi r^2dr$$
$$\sigma = \frac{M}{\frac{4}{3}\pi*R^3}$$
$$I = \sigma 4 \pi \int_0^R{r^4 dr} = \frac{3*MR^2}{5},$$
which is not the correct moment of inertia of a sphere
The liquid-vapor mass transfer (evaporation and condensation) is governed by the vapor transport equation:
$$\frac{\partial}{\partial t} (\alpha_l \rho) + \nabla \cdot (\alpha_l \rho \vec v) = \dot m^{+} + \dot m^{-}$$
In the incompressible flow case (constant density), it reduces to
$$...
Find the question here and the solution i.e number 10 indicated as ##6-4\sqrt{2}##,
I am getting a different solution, my approach is as follows. I made use of pythagoras theorem for the three right angle triangles as follows,
Let radius of the smaller circle be equal to ##c## and distance...
If there is a circle, 360 degrees, and 2 random points A & B are chosen on the circle. A straight line is drawn between those two points. You are not shown the whole circle. All you can see is A, B the straight line and an arc representing an unknown number of degrees. You know the length of the...
The result is supposed to be 12,2 m but every time I get 8,016 m... I used for example this formula >r=m/[(density of air-density of hot air)*(4/3)*pi]
For density I used > rho=(p*M)/(R*T)
Am I forgetting something? Thanks in advance.
I don't understand part (c) the answer is decreasing for both. I would have it decreasing for the centripetal acceleration and then increasing for the velocity, why is that not correct?
Thanks in advance!
My source for Ωk and H0 is
https://www.cosmos.esa.int/documents/387566/387653/Planck_2018_results_L06.pdf ,
page 40, equation 47b. Ωk = 0.0007.
Page 15 equation 13 has H0 = 67.27 km/(s Mpc).
The formula I used is
R = (c/H0) (1/Ωk)1/2 .
( I did have a reference for this, but I misplaced it, and I...
Greetings!
I have a problem with the solution of that exercice
I don´t agree with it because if i choose to factorise with 6^n instead of 2^n will get 5/6 instead thank you!
I have attempted to solve for the velocity by setting the centripetal force (mv2)/r to the normal force pointed to the center of rotation (mg). This approach seems to give the incorrect solution and I am unsure of my misunderstandings.
When I looked up for Bohr radius, the formula has ##q## in it, which is charge of the object. For this question, the electron and proton are replaced by sun and Earth so it means that I have to know the charge of Earth and sun?
Thanks
confused on part A/B when I look up they did E= Q/2e(0.8)^2.
But why not use the 0.100mm because that is the area of the enclosed.
Same with B why did they use 100m and not 0.8m because 0.8 is smaller so it enclosed the charge
I am able to prove that it is 0.225 but how do I prove that it is also 0.414?
I need to find the max. and min. packing fraction values, which I got as a function of (r1/r2)
Please help
I know that the whole topic of inductance in a straight wire is complicated (and has led to some heated discussions in this forum :smile:). I followed Rosa's derivation and can see that it leads to an inverse relation of the inductance to the wire radius, and from what could understand, the...
I used the above equation, and started with getting the cross product of dl and r, which was equal to 0.00195i+0.00365k. From there, I divided each component by the magnitude of radius cubed (0.827^3). I then multiplied by I and u naught(u_0=4pi*10^-7), and then divided by 4pi. The answer I got...
Greetings
I have some problems finding the correct result
My solution:
I puted Y=6x+16
so now will try to find the raduis of convergence of Y
so let's calculate the raduis criteria of convergence:
We know that Y=6x+16
Conseqyently -21/6<=x<=-11/6 so the raduis must be 5/3. But this is not...
Basically, I'm wondering if there have been any attempts to calculate/model what the radius of the sun should be based on gravitational, thermal, and electromagnetic pressures. If there has, where can I find the calculation/model, and how closely does it match the actual radius of the sun...
How did the author compute the highlighted term 2 from the highlighted term 1 in the following answer to the given question?
If $\rho =\frac{d\psi}{ds}$, then the term 2 should be $\upsilon^2 \frac{d\hat{T}}{d\psi}\rho$, but instead, it was written...
A steel pipe with foam insulation is embedded in a concrete wall. The steel pipe is carrying cold water and therefore gains heat from outside atmosphere. Heat transfers through the concrete wall, foam insulation and then to the pipe. Is it possible to calculate critical radius of insulation for...
Can someone please verify if my reasoning is accurate?
I chose E) Planets B and D because they both have the same ratio of mass to radius which is the lowest of all the other planet options. Due to the fact that they have mass and radius evened out the gravitational pull will pull weight down...