Answer:
The answer is "4-5-6 because the bank is growing too quickly in the beginning of its turn".
Explanation:
The S-turns is also known as the reference technique, in which the ground track of the aircraft on both sides of a defined ground-based straight-line distance represents 2 different but equivalent circles.
Throughout the S-turns, on either side of the road, a progressively smaller half-circle is formed, and this turn does not stop until the road or reference line is crossed during the early part of the turn, the twists increase too quickly.
Answer:
T = 0.03 Nm.
Explanation:
d = 1.5 in = 0.04 m
r = d/2 = 0.02 m
P = 56 kips = 56 x 6.89 = 386.11 MPa
σ = 42-ksi = 42 x 6.89 = 289.58 MPa
Torque = T =?
<u>Solution:</u>
σ = (P x r) / T
T = (P x r) / σ
T = (386.11 x 0.02) / 289.58
T = 0.03 Nm.
Answer:
a) 600nm
b) 300nm
Explanation:
the path difference = 2t
t = thickness of the film
L' = wavelength of light in film = L/n
L = wavength of light in air
n = refractive index of glass
(a)
for destructive interference 2t = L'/2 = L/2n
L = 4*t*n
= 4*120*10^-9*1.25
L = 600 nm
(b)
for constructive interference 2t = L' = L/1.25
L = 2tn
= 2 × 1.25 × 120nm
= 300 nm
A) mass m with F1 acting in the positive x direction and F2 acting perpendicular in the positive y direction<span>
m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N</span><span> ... y direction
Net force ^2 = F1^2 + F2^2 = (20N)^2 + (15n)^2 = 625N^2 =>
Net force = √625 = 25N
F = m*a => a = F/m = 25.0 N /5.00 kg = 5 m/s^2
Answer: 5.00 m/s^2
b) mass m with F1 acting in the positive x direction and F2 acting on the object at 60 degrees above the horizontal.
</span>
<span>m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N</span><span> ... 60 degress above x direction
Components of F2
F2,x = F2*cos(60) = 15N / 2 = 7.5N
F2, y = F2*sin(60) = 15N* 0.866 = 12.99 N ≈ 13 N
Total force in x = F1 + F2,x = 20.0 N + 7.5 N = 27.5 N
Total force in y = F2,y = 13.0 N
Net force^2 = (27.5N)^2 + (13.0N)^2 = 925.25 N^2 = Net force = √(925.25N^2) =
= 30.42N
a = F /m = 30.42 N / 5.00 kg = 6.08 m/s^2
Answer: 6.08 m/s^2
</span>
To solve this problem it is necessary to apply the concepts related to the magnetic dipole moment in terms of the current and the surface area, as well as the current density, as a function of the current over the area.
Part A) By definition we know that magnetic dipole moment is
Where,
I = Current
S = Area
Replacing with our values we have that,
Re-arrange to find I,
Part B) To find the Current density we need to find the cross sectional area of the Wire:
Finally the current density is simply J
PART C) Finally to make the comparison with the given values we have to cross-sectional area would be
Therefore the current density would be
Comparing the two values we can see that the 2mm wire has a higher current density.
