Answer:
the maximum frequency observed is 2.0044 10⁶ Hz
Explanation:
This is a Doppler effect exercise. Where the emitter is still and the receiver is mobile, therefore the expression that describes the process is
f ’= 
the + sign is used when the observer approaches the source
typical speeds of a baby's heart stop are around 200 m / min
let's reduce to SI units
v₀ = 200 m / min (1 min / 60 s) = 3.33 m / s
let's calculate
f ’= 2 10⁶ (
)
f ’= 2.0044 10⁶ Hz
f ’= 1,9956 10⁶ Hz
therefore the maximum frequency observed is 2.0044 10⁶ Hz
Dry air adjacent to the cloud is entrained air is drier than the air within a cloud. The evaporation occurs in the cloud which cools the air. The cooling of air increases its density and creates a downdraft.
<h3>How clouds are formed?</h3>
A cloud can be described as a mass of ice crystals or water drops suspended in the atmosphere. Clouds can be formed when the water condenses in the atmosphere. The sky possesses some quantity of water vapours and it is invisible to us.
Clouds can be formed when an area of air gets cooler until the water vapour there condenses to liquid form. At this point, the air gets saturated with water vapours.
A cloud can never be perfectly adiabatic. Therefore, after mixing the environmental air with the clouds, its boundaries will not stay well defined and this process is called entrainment.
Learn more about cloud formation, here:
brainly.com/question/1242352
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C. Echolocation
Echolocating animals emit calls out to the environment and listen to the echoes of those calls that return from various objects near them. They use these echoes to locate and identify the objects.
Answer:
Explanation:
There's an easy way to answer this and then an easier way. I'll do both since I'm not sure what you're doing this for: physics or calculus. Calculus is the easier way, btw.
Going with the physics version first, here's what we know:
a = -9.8 m/s/s
v₀ = 3.75 m/s
t = ??
That's not a whole lot...at least not enough to directly solve the problem. What we have to remember here is that at the max height of a parabolic path, the final velocity is 0. So we can add that to our info:
v = 0 m/s. Use the one-dimensional equation that utilizes all that info and allows us to solve for time:
v = v₀ +at and filling in:
0 = 3.75 + (-9.8)t and
-3.75 = -9.8t so
t = .38 seconds. This is how long it takes to get to its max height. Another thing we need to remember (which is why calculus is so much easier!) is that at the halfway point of a parabolic path (the max height), the object has traveled half the time it takes to make the whole trip. In other words, if .38 is how long it takes to go halfway, then 2(.38) is how long the whole trip takes:
2(.38) = .76 seconds. Now onto the calculus way:
The position function is
The first derivative of this is the velocity function and, knowing that when the velocity is 0, the time is halfway gone, we will find the velocity function and then set it equal to 0 and solve for t:
v(t) = -9.8t + 3.75 and
0 = -9.8t + 3.75 and
-3.75 = -9.8t so
t = ,38 and multiply that by 2 to find the time the whole trip took:
2(.38) = .76 seconds.
Answer:
Q1_new = 515.68 µC
Q2_new = 246.82 µC
Explanation:
Since the capacitors are charged in parallel and not in series, then both are at 250 V when they are disconnected from the battery.
Then it is only necessary to calculate the charge on each capacitor:
Q1 = 5.85 µF * 250 V = 1462.5 µC
Q2 = 2.8 µF * 250 V = 700 µC
Now, we will look at 1462.5 µC as excess negative charges on one plate, and 1462.5 µC as excess positive charges on the other plate. Now, we will use this same logic for the smaller capacitor.
When there is a connection of positive plate of C1 to the negative plate of C2, and also a connection of the negative plate of C1 to the positive plate of C2, some of these excess opposite charges will combine and cancel each other. The result is that of a net charge:
1462.5 µC - 700 µC = 762.5 µC
Thus,762.5 µC of net charge will remain in the 'new' positive and negative plates of the resulting capacitor system.
This 762.5 µC will be divided proportionately between the two capacitors.
Q1_new = 762.5 µC * (5.85/(5.85 + 2.8)) = 515.68 µC
Q2_new = 762.5 µC * (2.8/(5.85 + 2.8) = 246.82 µC
