Below are the choices that can be found elsewhere:
a. 268 kJ
<span>b. 271 kJ </span>
<span>c. 9 kJ </span>
<span>d. 6 kJ
</span>
So the key thing to realize here is what the information given to you actually means. Sublimation is going from a sold to a gas. Vaporization is going from a liquid to a gas. Hence you can create two equations from the information that you have:
<span>Ga (s) --> Ga (g) delta H = 277 kJ/mol </span>
<span>Ga (l) --> Ga (g) delta H = 271 kJ/mol </span>
<span>From these two equations, you can then infer how to get the melting equation be simply finding the difference between the sublimation (two steps) and vaporization (one step). </span>
<span>Ga (s) --> Ga (l) delta H = 6 kJ/mol </span>
<span>At this point, all you need to do is a bit of stoichiometry. You start with 1.50 mol and multiply by the amount of energy per mole (6 kJ/mol). </span>
<span>*ANSWER* </span>
<span>9 kJ/mol (C)</span>
W = 1/2k*x^2.
k = spring constant = 2500 n/m.
x = distance = 4 cm = 0.04m (convert to same units).
W = 1/2(2500)(0.04)^2 = 2J.
Answer:
Explanation:
a ) No of turns per metre
n = 450 / .35
= 1285.71
Magnetic field inside the solenoid
B = μ₀ n I
Where I is current
B = 4π x 10⁻⁷ x 1285.71 x 1.75
= 28.26 x 10⁻⁴ T
This is the uniform magnetic field inside the solenoid.
b )
Magnetic field around a very long wire at a distance d is given by the expression
B = ( μ₀ /4π ) X 2I / d
= 10⁻⁷ x 2 x ( 1.75 / .01 )
= .35 x 10⁻⁴ T
In the second case magnetic field is much less. It is due to the fact that in the solenoid magnetic field gets multiplied due to increase in the number of turns. In straight coil this does not happen .
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
