The gravitational force between <em>m₁</em> and <em>m₂</em> has magnitude

while the gravitational force between <em>m₁</em> and <em>m₃</em> has magnitude

where <em>x</em> is measured in m.
The mass <em>m₁</em> is attracted to <em>m₂</em> in one direction, and attracted to <em>m₃</em> in the opposite direction such that <em>m₁</em> in equilibrium. So by Newton's second law, we have

Solve for <em>x</em> :

The solution with the negative square root is negative, so we throw it out. The other is the one we want,

Answer:
d) False. If the angular momentum is zero, it implies in electro without turning, which would create a collapse towards the nucleus, so in both models the moment must be different from zero
Explanation:
Affirmations
a) true. The orbits are accurate in the Bohr model and probabilistic in quantum mechanics
b) True. If both give the same results and use the same quantum number (n)
c) True. If in angular momentum it is quantized, in the Bohr model too but it does not justify it
d) False. If the angular momentum is zero, it implies in electro without turning, which would create a collapse towards the nucleus, so in both models the moment must be different from zero
(I assume that the 4 directions north-south-east-west are meant with respect to the wire seen from the top.)
We can use the right-hand rule to understand the direction of the magnetic field generated by the wire. The thumb follows the direction of the current in the wire (upward), while the other fingers give the direction of the field in every point around the wire. Seen from the top, the field has an anti-clockwise direction. Therefore, if we take a point at east with respect to the wire, in this point the field has direction south.
Answer:
Wavelength is 471 nm
Explanation:
Given that,
Lines per unit length of diffraction grating is 500 lines/mm.
The third maxima from the central maxima (m=3) is at an angle of 45°
We need to find the color of laser light shines through a diffraction grating.
The condition for maxima is :

d = 1/N, N = number of lines per mm
