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
Answer:it helps get everyone in the same page
Explanation:
This is important so everyone knows what they should be doing
<span>The moment of inertia of the large sphere will be twice that of the smaller sphere.
The formula for the moment of inertia for a solid sphere is:
I = (2/5)mr^2
where
I = moment of inertia
m = mass
r = radius
Since both spheres have the same diameter, they also have the same radius, so the only change is their mass. And the moment of inertia is directly proportional to their mass as shown by the above formula. So the sphere with twice the mass will have twice the moment of inertia, or 2 times.</span>
<h3><u>Given</u> :</h3>
Current flow light bulb = 2.5
Resistance of light bulb = 3.6Ω
<h3><u>To Find </u>:</h3>
We have to find voltage of battery
<h3><u>Solution</u> :</h3>
➠ As per ohm's law, current flow through a conductor is directly proportional to the applied potential difference.
➝ V ∝ I
➝ <u>V = I × R</u>
Where, R is the resistance of conductor.
⇒ V = I × R
⇒ V = 2.5 × 3.6
⇒ <u>V = 9 volt</u>