The solution would be like
this for this specific problem:
<span>
The force on m is:</span>
<span>
GMm / x^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] ->
1
The force on 2m is:</span>
<span>
GM(2m) / (L - x)^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2]
-> 2
From (1), you’ll get M = 2mx^2 / L^2 and from
(2) you get M = m(L - x)^2 / L^2
Since the Ms are the same, then
2mx^2 / L^2 = m(L - x)^2 / L^2
2x^2 = (L - x)^2
xsqrt2 = L - x
x(1 + sqrt2) = L
x = L / (sqrt2 + 1) From here, we rationalize.
x = L(sqrt2 - 1) / (sqrt2 + 1)(sqrt2 - 1)
x = L(sqrt2 - 1) / (2 - 1)
x = L(sqrt2 - 1) </span>
= 0.414L
<span>Therefore, the third particle should be located the 0.414L x
axis so that the magnitude of the gravitational force on both particle 1 and
particle 2 doubles.</span>
The velocity vector of the planet points toward the center of the circle is the following is true about a planet orbiting a star in uniform circular motion.
A. The velocity vector of the planet points toward the center of the circle.
<u>Explanation:</u>
Motion of the planet around the star is mentioned to be uniform and around a circular path. Objects in uniform circular motion motion has constant angular speed but the velocity of the object will not remain constant. Since the planet is in circular motion the direction of velocity vector at a particular point is tangential to the circular path at that particular point.
Thus at every point, the direction of velocity vector changes and this means the velocity is never constant. The objects in uniform circular motion has centripetal acceleration which means that velocity vector of the planet points toward the center of the circle.
The force exerted on the board by the karate master given the data is -4500 N
<h3>Data obtained from the question </h3>
- Initial velocity (u) = 10 m/s
- Final velocity (v) = 1 m/s
- Time (t) = 0.002 s
- Mass (m) = 1 Kg
- Force (F) = ?
<h3>How to determine the force</h3>
The force exerted can be obtained as illustrated below:
F = m(v - u) / t
F = 1 (1 - 10) / 0.002
F = (1 × -9) / 0.002
F = -4500 N
Learn more about momentum:
brainly.com/question/250648
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Answer:
(a) 
(b) 
(c) 
(d) 
Solution:
As per the question:
Refractive index of medium 1, 
Angle of refraction for medium 1, 
Angle of refraction for medium 2, 
Now,
(a) The expression for the refractive index of medium 2 is given by using Snell's law:
where
= Refractive Index of medium 2
Now,
(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:
(c) To calculate the velocity of light in medium 1:
We know that:
Thus for medium 1
(d) To calculate the velocity of light in medium 2:
For medium 2:
