Aaron's car is moving at speed of 30 m/s
His reaction time is given as 0.7 s
but when he is tired the reaction time is doubled
Now we need to find the distance covered by his car when he is tired during the time when he react to apply brakes
So here since during this time speed is given as constant so we can say that distance covered can be product of speed and time
So here we can use



So the car will move to 42 m during the time when he apply brakes
Wouldn't it be neat if an electron falling closer to the nucleus ... emitting a
photon ... actually gave out more energy than it needed to climb to its original
energy level by absorbing a photon ! If there were some miraculous substance
that could do that, we'd have it made.
All we'd need is a pile of it in our basement, with a bright light bulb over the pile,
connected to a tiny hand-crank generator.
Whenever we wanted some energy, like for cooking or heating the house, we'd
switch the light bulb on, point it towards the pile, and give the little generator a
little shove. It wouldn't take much to git 'er going.
The atoms in the pile would absorb some photons, raising their electrons to higher
energy levels. Then the electrons would fall back down to lower energy levels,
releasing more energy than they needed to climb up. We could take that energy,
use some of it to keep the light bulb shining on the pile, and use the extra to heat
the house or run the dishwasher.
The energy an electron absorbs when it climbs to a higher energy level (forming
the atom's absorption spectrum) is precisely identical to the energy it emits when
it falls back to its original level (creating the atom's emission spectrum).
Energy that wasn't either there in the atom to begin with or else pumped
into it from somewhere can't be created there.
You get what you pay for, or, as my grandfather used to say, "For nothing
you get nothing."
Answer:
3.57 MJ
Explanation:
ASSUMING it's fresh water with density of 1000 kg/m³
W = ΔPE = mgΔh = 14.0(1000)(9.81)(26.0) = 3,570,840 J
Salt water would require more.
Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V
Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:
h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.
In order to calculate the cutoff wavelength we have to consider that Ek=0
in this case h*ν=W
(h*c)/λ=4.52 eV
λ= (h*c)/4.52 eV
λ= (1240 eV*nm)/(4.52 eV)=274.34 nm
From this h*ν = Ek+W; we can calculate the kinetic energy for a radiation wavelength of 198 nm
then we have
(h*c)/(λ)-W= Ek
Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV
Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this acts to slow down the ejected electrons from the catode.
The watt is a rate, similar to something like speed (miles per hour) and other time-interval related measurements.
Specifically, watt means Joules per Second. We are given that the electrical engine has 400 watts, meaning it can make 400 joules per second. If we need 300 kJ, or 3000 Joules, then we can write an equation to solve the time it would take to reach this amount of joules:
w * t = E
w: Watts
t: Time
E: Energy required
(Watts times time is equal to the energy required)
<u>Input our values:</u>
400 * t = 3000
(We need to write 3000 joules instead of 300 kilojoules, since Watts is in joules per second. It's important to make sure your units are consistent in your equations)
<u>Divide both sides by 400 to isolate t:</u>
<u />
= 
t = 7.5 (s)
<u>It will take 7.5 seconds for the 400 W engine to produce 300 kJ of work.</u>
<u></u>
If you have any questions on how I got to the answer, just ask!
- breezyツ