<span>Answer:
The temperature doesn't affect the evaporation rate, but affects on how much of water a parcel of air can contain when saturated which is known by the absolute humidity. Hurricanes are usually happening when the temperature of the sea water west of the Cape Verde islands is over 27 degrees Celsius. If ahead of the path of a hurricane, the sea water temperature drops then it will be less moisture in the air and perhaps the hurricane will fade out. But it is not as simple. How strong a tropical storm is is relative to the difference of temperture between ground level and the top of the troposphere. The greater the difference, the faster the air will rise and the deeper the pressure will be, forcing surrounding air to rush in, thus forming a hurricane force wind. Then there is the fact that the wet adiabatic lapse rate is about half that of dry air. It means that rising moist air cools down slower and therefore rises higher. Hence water is the true fuel of bad weather. But it can't be isolated from the fact that the difference of temperature must be great too. What we often forget is that the tropopause (the border to the stratosphere) is much higher over the equator and therefore, much colder than e.g. the poles.</span>
Explanation:
Given that,
Wavelength = 6.0 nm
de Broglie wavelength = 6.0 nm
(a). We need to calculate the energy of photon
Using formula of energy



(b). We need to calculate the kinetic energy of an electron
Using formula of kinetic energy


Put the value into the formula


(c). We need to calculate the energy of photon
Using formula of energy



(d). We need to calculate the kinetic energy of an electron
Using formula of kinetic energy


Put the value into the formula


Hence, This is the required solution.
The magnetic field at the center of the arc is 4 × 10^(-4) T.
To find the answer, we need to know about the magnetic field due to a circular arc.
<h3>What's the mathematical expression of magnetic field at the center of a circular arc?</h3>
- According to Biot savert's law, magnetic field at the center of a circular arc is
- B=(μ₀ I/4π)× (arc/radius²)
- As arc is given as angle × radius, so
B=( μ₀I/4π)×(angle/radius)
<h3>What will be the magnetic field at the center of a circular arc, if the arc has current 26.9 A, radius 0.6 cm and angle 0.9 radian?</h3>
B=(μ₀ I/4π)× (0.9/0.006)
= (10^(-7)× 26.9)× (0.9/0.006)
= 4 × 10^(-4) T
Thus, we can conclude that the magnitude of magnetic field at the center of the circular arc is 4 × 10^(-4) T.
Learn more about the magnetic field of a circular arc here:
brainly.com/question/15259752
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Explanation:
Light Light from a moving source travels at 300,000 km/sec (186,000 miles/sec).
Note:
The Speed of Light Is Constant.
Answer: current I = 1.875A
Explanation:
If the resistors are connected in series,
Then the equivalent resistance will be
R = 6 + 18 + 15 + 9
R = 48 ohms
Using ohms law
V = IR
Make current I the subject of formula
I = V/R
I = 90/48
I = 1.875A
And if the resistors are connected in parallel, the equivalent resistance will be
1/R = 1/6 + 1/18 + 1/15 + 1/9
1/R = 0.166 + 0.055 + 0.066 + 0.111
R = 1/0.3999
R = 2.5 ohms
Using ohms law
V = IR
I = 90/2.5
Current I = 35.99A