Answer:
0.44m/s
Explanation:
drift velocity=I/nAq
diameter 12 gauge
wire=0.081inches=0.081*2.5=0.2025cm radius=0.10125cm area=pi*R^2 =20/8.5*10^22*3.14*0.10125^2*10^-4*1.6*10^-19*
V = 0.44m/s
A solar eclipse occurs when the moon crosses in front of the Sun, blocking some or all of its rays. A lunar eclipse happens when the moon is directly behind the earth, blocking the moon from receiving light. The only light comes from the light on earth's reflected shadow.
You can look at a lunar eclipse because there is very little light or none at all. You can't look at a solar eclipse because you are looking directly at the sun unless it is complete. Before totality, only some of the Sun is blocked, causing your pupils dilate to let in more light. Since they do this, more of the Sun's rays can be let in to the eye, which effectively allows your eyes to burn.
Some doctors and eye care specialists say that after someone complains of blindness after looking at a solar eclipse unaided, they can see what the Sun and moon looked like at the time that they looked at it, as it is burned onto their retinas.
The so-called "terminal velocity" is the fastest that something can fall
through a fluid. Even though there's a constant force pulling it through,
the friction or resistance of plowing through the surrounding substance
gets bigger as the speed grows, so there's some speed where the resistance
is equal to the pulling force, and then the falling object can't go any faster.
A few examples:
-- the terminal velocity of a sky-diver falling through air,
-- the terminal velocity of a pecan falling through honey,
-- the terminal velocity of a stone falling through water.
It's not possible to say that "the terminal velocity is ----- miles per hour".
If any of these things changes, then the terminal velocity changes too:
-- weight of the falling object
-- shape of the object
-- surface texture (smoothness) of the object
-- density of the surrounding fluid
-- viscosity of the surrounding fluid .
The answer to the question is:
75m/s
Just do 25*3
Answer:
The latent heat of vaporization of water is 2.4 kJ/g
Explanation:
The given readings are;
The first (mass) balance reading (of the water) in grams, m₁ = 581 g
The second (mass) balance reading (of the water) in grams, m₂ = 526 g
The first joulemeter reading in kilojoules (kJ), Q₁ = 195 kJ
The second joulemeter reading in kilojoules (kJ), Q₂ = 327 kJ
The latent heat of vaporization = The heat required to evaporate a given mass water at constant temperature
Based on the measurements, we have;
The latent heat of vaporization = ΔQ/Δm
∴ The latent heat of vaporization of water = (327 kJ - 195 kJ)/(581 g - 526 g) = 2.4 kJ/g
The latent heat of vaporization of water = 2.4 kJ/g
