Answer:

Explanation:
y = Distance from the center point
d = Separation between slits = 0.3 mm
D = Distance between slit and screen = 1.5 m
= Wavelength = 700 nm
m = Order = 1
We have the relation

The distance from the screen at which the first bright fringe beyond the center fringe appear is
.
Answer:
The correct answer is:
a) remain where it is released
Explanation:
The concept of density seeks to measure the weight of an object in relation to its size. It is the measure of how packed together the particles of that object are. An object placed in a liquid displaces a certain volume of the liquid, based on the relative density of the object and the liquid.
If an object is less dense than a liquid in which it is placed, it displaces a smaller volume of the liquid than its volume, hence only some part of the object will be seen to be under the liquid, the other part will float.
If an object is denser than the liquid in which it is placed, it displaces a larger volume of the liquid than its own volume, making the object to sink and is submerged, sometimes to the bottom of the liquid, but mostly below the point at which it was released.
Finally, if the density of an object and the liquid into which it is submerged is the same. the object's mass per unit volume is the same as the liquid's mass per unit volume, hence the weight and force created due to density will balance and cancel each other out hence making the object to remain where it was submerged.
The answer is A. Hope this helps. :)
Answer:
a) ΔV = 25.59 V, b) ΔV = 25.59 V, c) v = 7 10⁴ m / s, v/c= 2.33 10⁻⁴ ,
v/c% = 2.33 10⁻²
Explanation:
a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy
starting point. Where the electrons come out
Em₀ = U = e DV
final point. Where they hit the target
Em_f = K = ½ m v2
energy is conserved
Em₀ = Em_f
e ΔV = ½ m v²
ΔV =
mv²/e (1)
If the speed of light is c and this is 100% then 1% is
v = 1% c = c / 100
v = 3 10⁸/100 = 3 10⁶6 m/ s
let's calculate
ΔV =
ΔV = 25.59 V
b) Ask for the potential difference for protons with the same kinetic energy as electrons
K_p = ½ m v_e²
K_p =
9.1 10⁻³¹ (3 10⁶)²
K_p = 40.95 10⁻¹⁹ J
we substitute in equation 1
ΔV = Kp / M
ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹
ΔV = 25.59 V
notice that these protons go much slower than electrons because their mass is greater
c) The speed of the protons is
e ΔV = ½ M v²
v² = 2 e ΔV / M
v² =
v² = 49,035 10⁸
v = 7 10⁴ m / s
Relation
v/c = 
v/c= 2.33 10⁻⁴