A man lifts various loads with the same lever. The distance of the applied force from the fulcrum is 2.00 m and the distance fro
2 answers:
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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
Let say for every 5 s of time interval the speed will remain constant
so it is given as
v(mi/h) 16 21 23 26 33 30 28
now we have to convert the speed into ft/s as it is given that 1 mi/h = 5280/3600 ft/s
so here we will have
v(ft/s) 23.5 30.8 33.73 38.13 48.4 44 41.1
now for each interval of 5 s we will have to find the distance cover for above interval of time
so here it will cover 1298.1 ft distance in 30 s interval of time