Which statement correctly describes the electrons in a water molecule?. . A.Which statement correctly describes the electrons in
2 answers:
Answer: Option (B) is the correct answer.
Explanation:
Molecular formula of water is . In a water molecule, hydrogen and oxygen ions are held together by strong hydrogen bonding.
Out of hydrogen and oxygen, oxygen is more electronegative. Thus, it will attract the electrons of hydrogen atom more towards itself. As a result, a partial positive charge will develop on hydrogen atom and a partial negative charge will develop on oxygen atom.
Thus, we can conclude that the statement electrons are pulled closer to the oxygen atom correctly describes the electrons in a water molecule.
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Answer:
a) the values of the angle α is 45.5°
b) the required magnitude of the vertical force, F is 41 lb
Explanation:
Applying the free equilibrium equation along x-direction
from the diagram
we say
∑Fₓ = 0
Pcosα - 425cos30° = 0
525cosα - 368.06 = 0
cosα = 368.06/525
cosα = 0.701
α = cos⁻¹ (0.701)
α = 45.5°
Also Applying the force equation of motion along y-direction
∑Fₓ = ma
Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)
525sin45.5° + F + 212.5 - 600 = 27.95
374.46 + F + 212.5 - 600 = 27.95
F - 13.04 = 27.95
F = 27.95 + 13.04
F = 40.99 ≈ 41 lb
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
Answer:
v = 61.09m/s
Explanation:
In order to calculate the speed of the electron when it is 3.00cm from the proton, you first calculate the acceleration of the electron, produced by the electric force between the electron and the proton. By using the second Newton law you have:
(1)
m: mass of the electron = 9.1*10^-31kg
q: charge of electron and proton = 1.6*10^-19C
r: distance between electron and proton = 9.00cm = 0.09m
k: Coulomb's constant = 8.98*10^9Nm2/C^2
You solve the equation (1) for a, and replace the values of the other parameters:
Next, you use the following formula to calculate the final speed of the electron:
(2)
vo: initial speed of the electron = 0m/s
a: acceleration = 3.11*10^4m/s^2
x: distance traveled by the electron
When the electron is at 3.00cm from the proton the electron has traveled a distance of 9.00cm - 3.00cm = 6.00cm = 0.06m = x
You replace the values of the parameters in the equation (2):
The speed of the electron is 61.09m/s