Answer:
NaNO3.
Explanation:
The Na ion has one positive charge and the NO3 ion has one negative charge so the correct formula is NaNO3.
Answer:
a) F = 3.2 10⁻¹⁰ N
, b) v = 9.9 10⁷ m / s
Explanation:
a) The electric force is
F = q E
The electric field is related to the potential reference
V = E d
E = V / d
Let's replace
F = e V / d
Let's calculate
F = 1.6 10⁻¹⁹ 28 10³ / 1.4 10⁻²
F = 3.2 10⁻¹⁰ N
b) For this part we can use kinematics
v² = v₀ + 2 a d
v = √ 2 ad
Acceleration can be found with Newton's second law
e V / d = m a
a = e / m V / d
a = 1.6 10⁻¹⁹ / 9.1 10⁻³¹ 28 10³ / 1.4 10⁻²
a = 3,516 10⁻¹⁷ m / s²
Let's calculate the speed
v = √ (2 3,516 10¹⁷ 1.4 10⁻²)
v = √ (98,448 10¹⁴)
v = 9.9 10⁷ m / s
Answer:
The magnetic field is strongest at the center and weakest between the two poles just outside the bar magnet. The magnetic field lines are densest at the center and least dense between the two poles just outside the bar magnet.
Explanation:
Answer:
the acceleration of harry is equal to 66.126 m/s²
Explanation:
given,
harry is 35 m behind Draco
speed of Draco = 40 m/s
original speed of harry = 50 m/s
acceleration = ?
time taken by the Draco
t =
t = 1.875 s
distance covered by Harry
d = 35 + 175 = 210 m
to calculate the acceleration of harry


a × 3.516 × 0.5 = 116.25
a = 66.126 m/s²
hence, the acceleration of harry is equal to 66.126 m/s²
Answer:
a
The number of fringe is z = 3 fringes
b
The ratio is 
Explanation:
a
From the question we are told that
The wavelength is 
The distance between the slit is 
The width of the slit is 
let z be the number of fringes that appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern is and this mathematically represented as

Substituting values
z = 3 fringes
b
From the question we are told that the order of the bright fringe is n = 3
Generally the intensity of a pattern is mathematically represented as
![I = I_o cos^2 [\frac{\pi d sin \theta}{\lambda} ][\frac{sin (\pi a sin \frac{\theta}{\lambda } )}{\pi a sin \frac{\theta}{\lambda} } ]](https://tex.z-dn.net/?f=I%20%3D%20I_o%20cos%5E2%20%5B%5Cfrac%7B%5Cpi%20d%20sin%20%5Ctheta%7D%7B%5Clambda%7D%20%5D%5B%5Cfrac%7Bsin%20%28%5Cpi%20a%20sin%20%5Cfrac%7B%5Ctheta%7D%7B%5Clambda%20%7D%20%29%7D%7B%5Cpi%20a%20sin%20%5Cfrac%7B%5Ctheta%7D%7B%5Clambda%7D%20%7D%20%5D)
Where
is the intensity of the central fringe
And Generally 
![I = I_o co^2 [ \frac{\pi (\frac{n \lambda}{d} )}{\lambda} ] [\frac{\frac{sin (\pi a (\frac{n \lambda}{d} ))}{\lambda} }{\frac{\pi a (\frac{n \lambda}{d} )}{\lambda} } ]](https://tex.z-dn.net/?f=I%20%3D%20I_o%20co%5E2%20%5B%20%5Cfrac%7B%5Cpi%20%28%5Cfrac%7Bn%20%5Clambda%7D%7Bd%7D%20%29%7D%7B%5Clambda%7D%20%5D%20%5B%5Cfrac%7B%5Cfrac%7Bsin%20%28%5Cpi%20a%20%28%5Cfrac%7Bn%20%5Clambda%7D%7Bd%7D%20%29%29%7D%7B%5Clambda%7D%20%7D%7B%5Cfrac%7B%5Cpi%20a%20%28%5Cfrac%7Bn%20%5Clambda%7D%7Bd%7D%20%29%7D%7B%5Clambda%7D%20%7D%20%5D)
![I = I_o cos^2 (n \pi)[\frac{\frac{sin(\pi a (\frac{n \lambda}{d} ))}{\lambda} )}{ \frac{ \pi a (\frac{n \lambda }{d} )}{\lambda} } ]](https://tex.z-dn.net/?f=I%20%3D%20I_o%20cos%5E2%20%28n%20%5Cpi%29%5B%5Cfrac%7B%5Cfrac%7Bsin%28%5Cpi%20a%20%28%5Cfrac%7Bn%20%5Clambda%7D%7Bd%7D%20%29%29%7D%7B%5Clambda%7D%20%29%7D%7B%20%5Cfrac%7B%20%5Cpi%20a%20%28%5Cfrac%7Bn%20%5Clambda%20%7D%7Bd%7D%20%29%7D%7B%5Clambda%7D%20%7D%20%5D)
![I = I_o cos^2 (3 \pi) [\frac{sin (\frac{3 \pi }{6} )}{\frac{3 \pi}{6} } ]](https://tex.z-dn.net/?f=I%20%3D%20I_o%20cos%5E2%20%283%20%5Cpi%29%20%5B%5Cfrac%7Bsin%20%28%5Cfrac%7B3%20%5Cpi%20%7D%7B6%7D%20%29%7D%7B%5Cfrac%7B3%20%5Cpi%7D%7B6%7D%20%7D%20%5D)

