I believe if it were heavier with more mass, then the sun would pull it in and there would be no mercury. It might also be hotter.
F = 130 revs/min = 130/60 revs/s = 13/6 revs/s
t = 31s
wi = 2πf = 2π × 13/6 = 13π/3 rads/s
wf = 0 rads/s = wi + at
a = -wi/t = -13π/3 × 1/31 = -13π/93 rads/s²
wf² - wi² = 2a∅
-169π²/9 rads²/s² = 2 × -13π/93 rads/s² × ∅
∅ = 1209π/18 rads
n = ∅/2π = (1209π/18)/(2π) = 1209/36 ≈ 33.5833 revolutions.
Answer:
a. I = 0.76 A
b. Z = 150.74
c. RL₁ = 34.41 , RL₂ = 602.58
d. RL₂ = 602.58
Explanation:
V₁ = 116 V , R₁ = 77.0 Ω , Vc = 364 V , Rc = 473 Ω
a.
Using law of Ohm
V = I * R
I = Vc / Rc = 364 V / 473 Ω
I = 0.76 A
b.
The impedance of the circuit in this case the resistance, capacitance and inductor
V = I * Z
Z = V / I
Z = 116 v / 0.76 A
Z = 150.74
c.
The reactance of the inductor can be find using
Z² = R² + (RL² - Rc²)
Solve to RL'
RL = Rc (+ / -) √ ( Z² - R²)
RL = 473 (+ / -) √ 150.74² 77.0²
RL = 473 (+ / -) (129.58)
RL₁ = 34.41 , RL₂ = 602.58
d.
The higher value have the less angular frequency
RL₂ = 602.58
ω = 1 / √L*C
ω = 1 / √ 602.58 * 473
f = 285.02 Hz
Answer:
Explanation:
Threshold frequency = 4.17 x 10¹⁴ Hz .
minimum energy required = hν where h is plank's constant and ν is frequency .
E = 6.6 x 10⁻³⁴ x 4.17 x 10¹⁴
= 27.52 x 10⁻²⁰ J .
wavelength of radiation falling = 245 x 10⁻⁹ m
Energy of this radiation = hc / λ
c is velocity of light and λ is wavelength of radiation .
= 6.6 x 10⁻³⁴ x 3 x 10⁸ / 245 x 10⁻⁹
= .08081 x 10⁻¹⁷ J
= 80.81 x 10⁻²⁰ J
kinetic energy of electrons ejected = energy of falling radiation - threshold energy
= 80.81 x 10⁻²⁰ - 27.52 x 10⁻²⁰
= 53.29 x 10⁻²⁰ J .
