Explanation:
The given data is as follows.
C = 
R = 
ohm
C
Q = 
Formula to calculate the time is as follows.
![Q_{t} = Q_{o} [e^{\frac{-t}{\tau}]](https://tex.z-dn.net/?f=Q_%7Bt%7D%20%20%3D%20Q_%7Bo%7D%20%5Be%5E%7B%5Cfrac%7B-t%7D%7B%5Ctau%7D%5D)
![13.5 \times 10^{-6} = 100 \times 10^{-6} [e^{\frac{-t}{2}}]](https://tex.z-dn.net/?f=13.5%20%5Ctimes%2010%5E%7B-6%7D%20%3D%20100%20%5Ctimes%2010%5E%7B-6%7D%20%5Be%5E%7B%5Cfrac%7B-t%7D%7B2%7D%7D%5D)
0.135 = 
= 7.407
t = 4.00 s
Therefore, we can conclude that time after the resistor is connected will the capacitor is 4.0 sec.
Answer:
4.47 km
Explanation:
If we draw the path of the van then we get a shape with two exposed points A and D. If we draw a line from point D perpendicular to BA we get point E. This gives us a right angled triangle ADE.
From Pythagoras theorem
AD² = AE² + ED²
Hence, the van is 4.47 km from its initial point
<span>
Reaction rates are affected by reactant concentrations and temperature. this is accounted for by the c</span>ollision model.
-Hope this helps.
Answer:
c) 2.02 x 10^16 nuclei
Explanation:
The isotope decay of an atom follows the equation:
ln[A] = -kt + ln[A]₀
<em>Where [A] is the amount of the isotope after time t, k is decay constant, [A]₀ is the initial amount of the isotope</em>
[A] = Our incognite
k is constant decay:
k = ln 2 / Half-life
k = ln 2 / 4.96 x 10^3 s
k = 1.40x10⁻⁴s⁻¹
t is time = 1.98 x 10^4 s
[A]₀ = 3.21 x 10^17 nuclei
ln[A] = -1.40x10⁻⁴s⁻¹*1.98 x 10^4 s + ln[3.21 x 10^17 nuclei]
ln[A] = 37.538
[A] = 2.01x10¹⁶ nuclei remain ≈
<h3>c) 2.02 x 10^16 nuclei</h3>
