All will have a dominant trait I can't see the following statments
Answer:
True b and c
Explanation:
In an RLC circuit the impedance is
![Z = \sqrt{[R^{2} + ( (wL)^{2} + (\frac{1}{wC})^{2} ] }](https://tex.z-dn.net/?f=Z%20%3D%20%5Csqrt%7B%5BR%5E%7B2%7D%20%2B%20%28%20%28wL%29%5E%7B2%7D%20%2B%20%28%5Cfrac%7B1%7D%7BwC%7D%29%5E%7B2%7D%20%5D%20%20%20%20%20%7D)
examine the different phrases..
a) False. The maximum impedance is the value of the resistance
b) True. Resonance occurs when
(wL)² + (1 / wC)² = 0
w² = 1 / LC
c) True. In resonance the impedance is the resistive part and the power is maximum
d) False. In resonance the inductive and capacitive part cancel each other out
e) False. The impedance is always greater outside of resonance, but at the resonance point they are equal
Answer:
It will take 15.55s for the police car to pass the SUV
Explanation:
We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:
1. ![x=x_{0}+vt](https://tex.z-dn.net/?f=x%3Dx_%7B0%7D%2Bvt)
2. ![x=x_{0}+v_{0}t+\frac{at^{2}}{2}](https://tex.z-dn.net/?f=x%3Dx_%7B0%7D%2Bv_%7B0%7Dt%2B%5Cfrac%7Bat%5E%7B2%7D%7D%7B2%7D)
Since both cars will travel the same distance x, we can equal both formulas and solve for t:
![vt = v_{0}t+\frac{at^2}{2}\\\\ 16\frac{m}{s}t =30\frac{m}{s}t-\frac{1.8\frac{m}{s^{2}} t^{2}}{2}](https://tex.z-dn.net/?f=vt%20%3D%20v_%7B0%7Dt%2B%5Cfrac%7Bat%5E2%7D%7B2%7D%5C%5C%5C%5C%20%20%2016%5Cfrac%7Bm%7D%7Bs%7Dt%20%3D30%5Cfrac%7Bm%7D%7Bs%7Dt-%5Cfrac%7B1.8%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%20t%5E%7B2%7D%7D%7B2%7D)
We simplify the fraction present and rearrange for our formula so that it equals 0:
![0.9\frac{m}{s^{2}} t^{2}-14\frac{m}{s}t=0 \\\\ t(0.9\frac{m}{s^{2}}t-14\frac{m}{s})=0](https://tex.z-dn.net/?f=0.9%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%20t%5E%7B2%7D-14%5Cfrac%7Bm%7D%7Bs%7Dt%3D0%20%5C%5C%5C%5C%20t%280.9%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7Dt-14%5Cfrac%7Bm%7D%7Bs%7D%29%3D0)
In the very last step we factored a common factor t. There is two possible solutions to the equation at
and:
![0.9\frac{m}{s^{2}}t-14\frac{m}{s}=0 \\\\ 0.9\frac{m}{s^{2}}t =14\frac{m}{s} \\\\ t =\frac{14\frac{m}{s}}{0.9\frac{m}{s^{2}}}=15.56s](https://tex.z-dn.net/?f=0.9%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7Dt-14%5Cfrac%7Bm%7D%7Bs%7D%3D0%20%5C%5C%5C%5C%20%200.9%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7Dt%20%3D14%5Cfrac%7Bm%7D%7Bs%7D%20%5C%5C%5C%5C%20t%20%3D%5Cfrac%7B14%5Cfrac%7Bm%7D%7Bs%7D%7D%7B0.9%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%7D%3D15.56s)
What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at
(when the SUV passed the police car) and
(when the police car catches up to the SUV)
30 minutes I am not sure about that
Answer:
10. 36 g ZnCl2
Explanation:
Zn + 2HCl -> ZnCl2 + H2
0.076 mol Zn
1.37 mol HCl
3 mol H2
Limiting reactant: Zn
1 mol Zn -> 1 mol ZnCl2
0.076 mol Zn ->x x= 0.076 mol ZnCl2=10.36 g