With gravitational acceleration at 9.8, initial height at 3.5m and distance at 22m the initial horizontal velocity is 26.03 ms and the flight time is .845 seconds
Answer: His average speed in mph over the last 400 m is 7.7 m/s.
Explanation:
Given: Hicham El Guerrouj of Morocco holds the world record in the 1500 m running race. He ran the final 400 m in a time of 51.9 s.
We know that , speed = 
Here , distance = 400m and time = 51.9 s
Then, speed = 
Hence, his average speed in mph over the last 400 m is 7.7 m/s.
The two flaws in
her experiment’s design are
<span>- She introduced at least one confounding variable.</span>
<span>- She tried to test multiple hypotheses at a time</span>
In the above mentioned experiment she had to have four samples to prove
four hypotheses, each one separately and not to mix two hypotheses in an alone
sample, that what it brings as consequence is the confusion.
<span>Depends on the precision you're working to.
proton mass ~ 1.00728 amu
neutron mass ~ 1.00866 amu
electron mass ~ electron mass = 0.000549 amu
Binding mass is:
mass of constituents - mass of atom
Eg for nitrogen:
(7*1.00728)-(7*1.00866)-(7*0.000549)
-14.003074 = 0.11235amu
Binding energy is:
E=mc^2 where c is the speed of light. Nuclear physics is usually done in MeV[1] where 1 amu is about 931.5MeV/c^2. So:
0.11235 * 931.5 = 104.6MeV
Binding energy per nucleon is total energy divided by number of nucleons. 104.6/14 = 7.47MeV
This is probably about right; it sounds like the right size!
Do the same thing for D/E/F and recheck using your numbers & you shouldn't go far wrong :)
1 - have you done this? MeV is Mega electron Volts, where one electronVolt (or eV) is the change in potential energy by moving one electron up a 1 volt potential. ie energy = charge * potential, so 1eV is about 1.6x10^-19J (the same number as the charge of an electron but in Joules).
It's a measure of energy, but by E=mc^2 you can swap between energy and mass using the c^2 factor. Most nuclear physicists report mass in units of MeV/c^2 - so you know that its rest mass energy is that number in MeV.</span>
Answer:
98.15 lb
Explanation:
weight of plane (W) = 5,000 lb
velocity (v) = 200 m/h =200 x 88/60 = 293.3 ft/s
wing area (A) = 200 ft^{2}
aspect ratio (AR) = 8.5
Oswald efficiency factor (E) = 0.93
density of air (ρ) = 1.225 kg/m^{3} = 0.002377 slugs/ft^{3}
Drag = 0.5 x ρ x 
x A x Cd
we need to get the drag coefficient (Cd) before we can solve for the drag
Drag coefficient (Cd) = induced drag coefficient (Cdi) + drag coefficient at zero lift (Cdo)
where
- induced drag coefficient (Cdi) =
(take note that π is shown as n and ρ is shown as 
)
where lift coefficient (Cl)= 
=
= 0.245
therefore
induced drag coefficient (Cdi) = = 
= 0.0024
- since the airplane flies at maximum L/D ratio, minimum lift is required and hence induced drag coefficient (Cdi) = drag coefficient at zero lift (Cdo)
- Cd = 0.0024 + 0.0024 = 0.0048
Now that we have the coefficient of drag (Cd) we can substitute it into the formula for drag.
Drag = 0.5 x ρ x x A x Cd
Drag = 0.5 x 0.002377 x (293.3 x 293.3) x 200 x 0.0048 = 98.15 lb
