ELECTROSTATIC:
relating to stationary electric charges or fields as opposed to electric currents.
NEUTRAL:
nor negative nor positive/having no charge
POSITIVELY CHARGED:
positive charge occurs when the number of protons exceeds the number of electrons
NEGATIVELY CHARGED:
negative charge occurs when the number of electrons exceeds the number of protons.
COULOMB:
SI unit for electric charge. One coulomb is equal to the amount of charge from a current of one ampere flowing for one second.
MICROCOULOMB:
a unit of electrical charge equal to one millionth of a coulomb.
NANOCOULOMB:
Nanocoulombs are a unit of charge 1,000,000,000 times smaller than Coulomb.
CONSERVATION OF CHARGE:
constancy of the total electric charge in the universe or in any specific chemical or nuclear reaction
QUANTISATION OF CHARGE:
Charge quantization is the principle that the charge of any object is an integer multiple of the elementary charge.
It has a 10 electrons. Since it's atomic number is 11 it must have 11 protons. Also, given that it has a +1 charge, it has one less electron than protons since they have equal but opposite charges.
The number of protons is the mass number minus the atomic number = 23-11= 12 neutrons.
a) At a position of 2.0m, the Initial energy is
all made up of the potential energy=m*g*hi<span>
and meanwhile at 1.5 all its energy is also potential energy=m*g*hf
The percentage of energy remaining is E=m*g*hi/m*g*hf x 100
and since mass and gravity are constant so it leaves us with
just E=hi/hf
which 1.5/2.0 x100= 75% so we see that we lost 25% of the
energy or 0.25 in fraction
b) Here use the equation vf^2=vi^2+2gd
<span>where g is gravity, vf is the final velocity and vi is the
initial velocity while d is the distance travelled
so in here we are looking for the vi so let us isolate that
variable
we know that at maximum height or peak, the velocity is 0 so
vf is 0
therefore,</span></span>
vi =sqrt(-2gd) <span>
vi =sqrt(-2x-9.81x1.5) </span>
<span>vi =5.4 m/s
<span>c) The energy was converted to heat due to friction with the
air and the ground.</span></span>
Answer:
a. 2v₀/a b. 2v₀/a
Explanation:
a. Since you are moving with a constant velocity v₀, the distance, s you cover in time = t max is s = v₀t.
Since the dragster starts from rest with an acceleration, a, using
s' = ut + 1/2at² where u = 0 and s' = distance moved by dragster
s' = 0t + 1/2at²
s' = 1/2at²
Since the distance moved by me and the dragster must be the same,
s = s'
v₀t. = 1/2at²
v₀t. - 1/2at² = 0
t(v₀ - 1/2at) = 0
t= 0 or v₀ - 1/2at = 0
t= 0 or v₀ = 1/2at
t= 0 or t = 2v₀/a
So the maximum time tmax = 2v₀/a
b. Since the distance covered by me to meet the dragster is s = v₀t in time, t = tmax which is also my distance from the dragster when it started. So, my distance from the dragster when it started is s = v₀(2v₀/a)
= 2v₀/a
