Answer:
230.4 N
Explanation:
From the question given above, the following data were obtained:
Charge (q) of each protons = 1.6×10¯¹⁹ C
Distance apart (r) = 1×10¯¹⁵ m
Force (F) =?
NOTE: Electric constant (K) = 9×10⁹ Nm²/C²
The force exerted can be obtained as follow:
F = Kq₁q₂ / r²
F = 9×10⁹ × (1.6×10¯¹⁹)² / (1×10¯¹⁵)²
F = 9×10⁹ × 2.56×10¯³⁸ / 1×10¯³⁰
F = 2.304×10¯²⁸ / 1×10¯³⁰
F = 230.4 N
Therefore, the force exerted is 230.4 N
Answer:
266 g or 0.266 kg
Explanation:
The formula for specific heat capacity is given as,
Q = cm(t₂-t₁) ..................... Equation 1
Where Q = Heat Energy, c = specific heat capacity of Aluminum, m = mass of the aluminum fins, t₁ = initial temperature, t₂ = final temperature.
make m the subject of the equation,
m = Q/c(t₂-t₁)................... Equation 2
Given : Q = 2571 J, c = 0.897 J/g.°C, t₁ = 15.73 °C, t₂ = 26.50 °C.
Substitute into equation 2
m = 2571/[0.897×(26.5-15.73)]
m = 2571/9.661
m = 266 g or 0.266 kg
Hence the mass of the Aluminum fins = 266 g or 0.266 kg
Answer:
1 day
Explanation:
Let the safe level = x
The current level = x + 0.2x = 1.2 x
Thus,
Half life = 3.8 days
Where, k is rate constant
So,
The rate constant, k = 0.1824 days⁻¹
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
So,
![\frac {[A_t]}{[A_0]}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D)
= x / 1.2 x = 0.8333
t = ?
![\frac {[A_t]}{[A_0]}=e^{-k\times t}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D%3De%5E%7B-k%5Ctimes%20t%7D)
t ≅ 1 day
<u>Lab must be vacated in 1 day.</u>
