Answer:
500.3 Bq
Explanation:
From the formula;
0.693/t1/2 = 2.303/t log (Ao/A)
Where;
t1/12 = half life = 5 years
Ao = initial activity = 2000 Bq
0.693/5 = 2.303/10 log (2000/A)
0.1386= 0.2303 log (2000/A)
0.1386/0.2303 = log (2000/A)
0.6018 = log (2000/A)
(2000/A) = Antilog (0.6018)
(2000/A) = 3.9976
A = 2000/3.9976
A = 500.3 Bq
Answer:
OK + MgBR arrow KBR + MG
Explanation:
I know nothing about this topic, but if it has to be balanced I am pretty certain that's the only balanced equation
Answer:
the SI base unit of electrical current
Coulomb's Law
Given:
F = 3.0 x 10^-3 Newton
d = 6.0 x 10^2 meters
Q1 = 3.3x 10^-8 Coulombs
k = 9.0 x 10^9 Newton*m^2/Coulombs^2
Required:
Q2 =?
Formula:
F = k • Q1 • Q2 / d²
Solution:
So, to solve for Q2
Q2 = F • d²/ k • Q1
Q2 = (3.0 x 10^-3 Newton) • (6.0 x 10^2 m)² / (9.0 x 10^9
Newton*m²/Coulombs²) • (3.3x 10^-8 Coulombs)
Q2 = (3.0 x 10^-3 Newton) • (360 000 m²) / (297 Newton*m²/Coulombs)
Q2 = 1080 Newton*m²/ (297 Newton*m²/Coulombs)
Then, take the reciprocal of the denominator and start
multiplying
Q2 = 1080 • 1 Coulombs/297
Q2 = 1080 Coulombs / 297
Q2 = 3.63636363636 Coulombs
Q2 = 3.64 Coulumbs
Answer:
Explanation: Bagong bayani ang turing sa ating mga Overseas Filipino. Workers (OFWs) sapagkat ang kanilang pagpapagod at pakikipagsapalaran sa ibayong dagat ay hindi lamang ... kasigurahan ang pagkukunan ni Lito sa panahon ng kanyang.