a new planet is found with a density one third as much at earth and a radius twice that of earth. what is the acceleration due t
1 answer:
You might be interested in
Answer:
a) t = 3.2 10¹⁴ s , b) Δm = 1,456 10⁻¹³ kg, mass decreases
Explanation:
In this exercise the removed electrons leave a positive charge on each disk, as the two charges are of the same sign they repel each other, so using Newton's equilibrium equation
–W = 0
F_{e} = W = mg
Electric force is
F_{e} = k q₁ q₂ / r²
Since the two generators remove 500 e/s, q₁ = q₂ = q
F_{e} = k q² / r²
k q² / r² = m g
q = √ m g r²/ k
q = √ (60 9.8 0.001² / 8.99 10⁹)
q = √ 6.54 10⁻⁴
q = 2.56 10⁻² C
This is the charge on each disk
q = # _electron e
# _electron = q / e
# _electron = 2.56 10⁻² / 1.6 10⁻¹⁹
# _electron = 1.6 10¹⁷ electrons
Let's use a rule of proportions to find the accumulation time, if the rate is 500 e/s
t = 1.6 10¹⁷ 1/500
t = 3.2 10¹⁴ s
b) how electrons are losing their mass decreases
Δm = me # _electron
Δm = 9.1 10⁻³¹ 1.6 10¹⁷
Δm = 1,456 10⁻¹³ kg