If we're talking about part c then
igneous rock + weathering and erosion = sedimentary rock
sedimentary rock + heat and pressure = metamorphic rock
magma + cooling = igneous rock
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:
(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:
4.86 m
Explanation:
Given that,
The frequency produced by a humming bird, f = 70 Hz
The speed of sound, v = 340 m/s
We need to find how far does the sound travel between wing flaps. Let the distance is equal to its wavelength. So,

So, the sound travel 4.86 m between wings flaps.
Answer:
d' = d /2
Explanation:
Given that
Distance = d
Voltage =V
We know that energy in capacitor given as



If energy become double U' = 2 U then d'



2 d ' = d
d' = d /2
So the distance between plates will be half on initial distance.
<u>Answer:</u>
950-kg subcompact car be moving to have the same kinetic energy of truck at 116.08 km/hr
<u>Explanation:</u>
Kinetic energy of body with mass m and moving at velocity v is given by

KE of truck 
KE of subcompact car 
Equating

So 950-kg subcompact car be moving to have the same kinetic energy of truck at 116.08 km/hr