That is false I think sorry if its not correct
Pt-200 is most likely to undergo beta-minus decay. Two four given isotopes of platinum are radioactive. Pt-192 is a stable isotope.
Answer:
t = 1.098*RC
Explanation:
In order to calculate the time that the capacitor takes to reach 2/3 of its maximum charge, you use the following formula for the charge of the capacitor:
![Q=Q_{max}[1-e^{-\frac{t}{RC}}]](https://tex.z-dn.net/?f=Q%3DQ_%7Bmax%7D%5B1-e%5E%7B-%5Cfrac%7Bt%7D%7BRC%7D%7D%5D)
(1)
Qmax: maximum charge capacity of the capacitor
t: time
R: resistor of the circuit
C: capacitance of the circuit
When the capacitor has 2/3 of its maximum charge, you have that
Q=(2/3)Qmax
You replace the previous expression for Q in the equation (1), and use properties of logarithms to solve for t:
![Q=\frac{2}{3}Q_{max}=Q_{max}[1-e^{-\frac{t}{RC}}]\\\\\frac{2}{3}=1-e^{-\frac{t}{RC}}\\\\e^{-\frac{t}{RC}}=\frac{1}{3}\\\\-\frac{t}{RC}=ln(\frac{1}{3})\\\\t=-RCln(\frac{1}{3})=1.098RC](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B2%7D%7B3%7DQ_%7Bmax%7D%3DQ_%7Bmax%7D%5B1-e%5E%7B-%5Cfrac%7Bt%7D%7BRC%7D%7D%5D%5C%5C%5C%5C%5Cfrac%7B2%7D%7B3%7D%3D1-e%5E%7B-%5Cfrac%7Bt%7D%7BRC%7D%7D%5C%5C%5C%5Ce%5E%7B-%5Cfrac%7Bt%7D%7BRC%7D%7D%3D%5Cfrac%7B1%7D%7B3%7D%5C%5C%5C%5C-%5Cfrac%7Bt%7D%7BRC%7D%3Dln%28%5Cfrac%7B1%7D%7B3%7D%29%5C%5C%5C%5Ct%3D-RCln%28%5Cfrac%7B1%7D%7B3%7D%29%3D1.098RC)
The charge in the capacitor reaches 2/3 of its maximum charge in a time equal to 1.098RC
Answer:
The answer to your question is given below
Explanation:
Since both object A and B were dropped from the same height and the air resistance is negligible, both object A and B will get to the ground at the same time.
From the question, we were told that object A falls through a distance to dA at time t and object B falls through a distance of dB at time 2t.
Remember, both objects must get to the ground at the same time..!
Let the time taken for both objects to get to the ground be t.
Time A = Time B = t
But B falls through time 2t
Therefore,
Time A = Time B = 2t
Height = 1/2gt^2
For A:
Time = 2t
dA = 1/2 x g x (2t)^2
dA = 1/2g x 4t^2
For B
Time = t
dB = 1/2 x g x t^2
Equating dA and dB
dA = dB
1/2g x 4t^2 = 1/2 x g x t^2
Cancel out 1/2, g and t^2
4 = 1
4dA = dB
Divide both side by 4
dA = 1/4 dB
Answer:
14.2
Explanation:
find horizontal force of the weight = 2.5kg x 9.8 Sin30 = 12.3 N
to prevent the sliding she needs to pull horizontally
Fh = 12.3 /Cos 30 =14.2N
