What happens if the distance between objects is 0 in Newton's law of universal gravitation?
A similar question was asked earlier today:How does physics deal with infinities?And since my answer for that question perfectly answers this, I will shamelessly copy it here (with a few modifications):Newton's law of gravitation only deals with point masses (i.e, masses that take up no volume). Since masses clearly do take up space in the physical world, we have to account for this. If the masses used in Newtons Law are treated as infinitesimally small and we integrate the force law over the volume/object the infinitessimal masses are contained in, we can calculate the force due to that object on other masses.When in this context it is easier to define something called the gravitational potential. The gravitational potential is a function of position described by [math]\nabla^2 V(\vec r)[/math][math]=4\pi G\rho(\vec r)[/math], where [math]\rho(\vec r)[/math] is the mass-density as a function of position; you can determine the gravitational force on some mass [math]m[/math] by the equation [math]\vec F=-m\vec\nabla V[/math]. This formulation of gravity is mathematically equivalent to Newton's Law of Gravitation.If your density is simply described by a delta function [math]\rho(\vec r)=m\delta(\vec r)[/math] (i.e, you are looking at a point mass), then you recover the inverse-square law, and your force will grow to an infinite magnitude as you get close to it. However, most of the time the density is treated as finite and spread out over a volume; here, you avoid the infinities because a finite mass is not confined to an infinitely small volume. If the density is spread out uniformly over a sphere's volume, for example, the gravitational force will go as a function of [math]M\frac{1}{r^2}[/math], where [math]M[/math] is the total mass of the sphere, for any radius outside of the spheres volume, but not once you are actually inside the sphere, and you avoid the problem of an infinite force.The conclusion I am trying to make is that infinities are avoided because mass is not confined to infinitely small points.
What is the distance between God and yourself?
It depends on your definition of God.Do you believe in an Infinite God?Do you believe in a God who is present everywhere and absent nowhere?If the answer to the above two questions is yes, then there is no distance between God and yourself.We are eternal individuations of an infinite field of intelligence.We are eternal points of view of the collective view of God.We are aspects of this infinite and eternal of intelligence which is the source and the substance of all that is - both perceived and not perceived.When you are one with anything, the only separation is in your thoughts, not in fact.While we are one with Source, we also have a personal relationship with Source. It is the eternal collective self of which we are eternal individuated selves. This is the subject for another post.If we believe in a God who is separate from us, separate from his creation, then even though God is the core and the ground of our being, we will always feel separated from God. And any feeling of separation of God and us might as well be a galactic distance since an inch and a million light-years is still separation.
If a body is travelling 7/16 of the total distance in the last second, what is the height of its fall?
Can it be assumed that the ball is dropped , so then u=0the total time to fall h metres is t secs , so it falls [9/16].h in [t-1] secss = ut +[1/2] gt^2 if u=0 , thenh = [1/2]gt^2 Eq 19/16]h=1/2g[t-1]^2 , Eq 2 , divide Eq 1 by Eq 2 , cancel h , and [1/2]g invert 9/1616/9 = t^2/[t-1]^216[t-1]^2 =9t^27t^2 -32t +16 =0[7t-4][t-4]=0t[1] = 4/7 is not a feasible solution , sot-4 secs is the time taken to fall h metresh= [1/2]g t^2 = 4.9[4^2] = 78.4 m
What is the distance between the points A (sin-cos, 0) and B (0, sin+cos)?
Answer: √2Solution:If the x-y axes are perpendicular, the (shortest) distance d between two points A and B having co-ordinates (x₁, y₁) and (x₂, y₂) respectively in Euclidean plane is given by the formulad = +√[(x₁ - x₂)² + (y₁ - y₂)²] …………………………………………………..(1)Given, A = (sin θ - cos θ, 0) and B = (0, sin θ + cos θ)∴ x₁ = sin θ - cos θ, y₁ = 0and x₂ = 0, y₂ = sin θ + cos θSubstituting these values of x₁, x₂ and y₁, y₂ in equation (1),d = √[( sin θ - cos θ - 0)² + (0 - sin θ - cos θ)²]= √[( sin θ - cos θ)² + (-1)²(sin θ + cos θ)²]= √[( sin θ - cos θ)² + (sin θ + cos θ)²]= √( sin² θ - 2sin θ cos θ + cos² θ + sin² θ + 2sin θ cos θ + cos² θ)= √[( sin² θ + cos² θ) + (sin² θ + cos² θ)] (Cancelling the term 2sin θ cos θ)= √( 1 + 1) = √2Hence the distance between the points A=(sin θ-cos θ,0) & B=(0, sin θ+cos θ)= √2 (Answer)
How do I prove that a disk charge resembles a point charge if its radius shrinks to 0?
Draw a line through the centre of the disc, of radius r, at right angles to it, and find the electric field at a point P, on this line, and a distance d from the disc.If the charge on the disc is Q, then its surface charge density is Q/πr². Consider a circular ring of the disc, of radius s and thickness ds. Work out the vertical electric field produced by this disc, and then integrate the formula, with respect to s, for s going from 0 to r.Letting r tend to zero will produce the formula E = k Q/d², which is the formula for a point charge Q, at a distance d from the disc.
In which situation is the displacement of a body equal to its distance covered?
As we know the displacement is the shortest distance between the initial and final positions.And the distance is the length of the actual path.If the object travels in the straight path the length of the actual path and the shortest distance are same. So the distance and the displacement are equal to each other if the object travels in the straight path
Eight solid uniform cubes of edge l are stacked together to form a single cube with the center o. One cube is removed from the system. What is the distance of the centre of mass of the remaining 7 cubes from o ?
The CM of the big cube is at O (0,0,0) and the CM of the picked out cube w.r.t. O is at (1/2,1/2,1/2) because the cube has side length 1.Let the CM of remaining system be at (X,Y,Z).By formulae of negative mass-X=((0*8m)-(1/2*m))/(8m-m)... where m is mass of each uniform cube.X=(-1/2)/7=-1/14By 3D symmetry we can say, X=Y=Z=(-1/14)Thus, CM of remaining system is at (-1/14,-1/14,-1/14)N.B. for any mistake in calculation please inform.
At what distance is the neutral point when a test charge is present in between +Q and -Q?
Consider a point P as neutal point in a system containing the charges +Q and -Q. Let d be the distance between them. Since the system has two charges which are equal in magnitude and opposite in sign, let us take it as a dipole. You know that the distance between a dipole is so small. So take d as small. The neutral point is the point where the force experienced by a unit positive charge is zero. So at point p, the +1ve charge experiences a force by both +Q and -Q. But since it is a neutral point, the net force is zero.Now lets go to the problem.Hope this gives you an answer for your question. If so.ething is wrong in my assumption please make me know it. So that i can have a right knowledge. If there is any mistake let me know it.
The line y=2.9x intersects the circle x^2 + y^2 - 200x - 4400 =0 at two points. What is the distance between the two points?
“The line y=2.9x intersects the circle x^2 + y^2 - 200x - 4400 =0 at two points. What is the distance between the two points?”Let the points be [math](x_1, y_1)[/math] and [math](x_2, y_2)[/math].The distance is then [math]d = \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2}[/math], or[math]d = \sqrt {(x_2-x_1)^2 + 2.9^2(x_2-x_1)^2}[/math][math]d = |x_2-x_1| \times \sqrt {1 + 2.9^2}[/math][math]d = |x_2-x_1| \times \sqrt {9.41}[/math]Let’s determine the x coordinates of the 2 points:[math]x^2+(2.9x)^2–200x-4400=0[/math][math]9.41x^2–200x-4400=0[/math][math]x=\frac {200 \pm \sqrt {40000+4 \times 9.41 \times 4400}} {18.82}[/math][math]x=\frac {200 \pm 4\sqrt {12851}} {18.82}[/math][math]x_1=\frac {200 + 4\sqrt {12851}} {18.82}[/math][math]x_2=\frac {200 - 4\sqrt {12851}} {18.82}[/math][math]|x_2 - x_1| = \frac {8\sqrt {12851}} {18.82}[/math]Replacing above:[math]d = \frac {8\sqrt {12851}} {18.82} \times \sqrt {9.41} = \frac {8\sqrt {12851}} {2 \sqrt {9.41}} = 4 \sqrt {\frac {12851} {9.41}} = 40 \sqrt {\frac {12851} {941}}[/math][math]d = 147.82…[/math]
Can you fnd the midpoint between the two points (-19, -14) and (3, 0)?
Finding a mid point between two coordinates is simple. Mid-point literally means halfway coordinates of X- and Y- axes. Therefore, we adopt the following technique:x_1 = -19, x_2 = 3 => x= (x_1+x_2)/2 => (-19+3)/2 = -8y_1 = -19, y_2 = 3 => y= (y_1+y_2)/2 => (-14+0)/2 = -7Therefore, the mid-point coordinates are (-8,-7).I hope this answer helps.