A capacitor stores 5.6 x 10-7 C of charge when connected to a 6.0-V battery. How much charge does the capacitor store when conne
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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
(a) The launching velocity of the beetle is 6.4 m/s
(b) The time taken to achieve the speed for launch is 1.63 ms
(c) The beetle reaches a height of 2.1 m.
(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.
Use the equation of motion,
Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.
Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.
The launching speed of the beetle is <u>6.4 m/s</u>.
(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,
Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.
The time taken by the beetle to launch itself upwards is <u>1.62 ms</u>.
(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.
Use the equation of motion,
Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.
The beetle can jump to a height of <u>2.1 m</u>