Answer:
F = 19.375 x 10^-6 N
Explanation:
This problem can be solved by applying Coulomb's Law, which lets us determine the force between two electrically charged particles.
It is defined as
F = (ke * q1 * q2)/ r^2
Where,
ke = is Coulomb's constant ≈ 9×10^9 N⋅m^2⋅C^−2
q1 = 5.0 x 10^-8 C
q2 = 1.0 x 10^-7 C
r = 5 ft = 1,524 m
F = (9×10^9 N⋅m^2⋅C^−2)*(5.0 x 10^-8 C)*(1.0 x 10^-7 C)/ ((1,524 m)^2)
F = (9×10^9 N⋅m^2⋅C^−2)*(5.0 x 10^-8 C)*(1.0 x 10^-7 C)/ ((1,524 m)^2)
F = 19.375 x 10^-6 N
<span>In assumption that there were two scientists who used
different measurement systems in their research. The problems that might arise
if they shared their data is obviously and primarily error. Errors are
recognized and one element in every measure, system and quantity. Error was
already even present in the measurement system alone a scientist used and it
will furthermore aggregate, when these two different scales are combined the
more error escalates in the process. There are two types: random and systematic
error. </span>
True true true truer than the
24 knots is the maximum wind velocity for a 30 crosswind if the maximum crosswind component for the airplane is 12 knots.
What is crosswind component?
Although you should always prepare to land the aircraft into the wind, in practice the wind only occasionally blows directly down the runway. You will typically be landing with at least a small amount of crosswind. There is a limit to how much direct crosswind an airplane can land in. Only a portion of the wind is operating as a direct crosswind when it is blowing at a shallow (acute) angle to the runway.
The following are the variables needed to solve for the crosswind component:
Runway and wind direction
Wind speed
To learn more about crosswind component click the given link
brainly.com/question/27960498
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Answer:
The answer is below
Explanation:
Let vₐ be the speed of airplane = 135 mph, vₙ be the speed of the wind = 70 mph and vₐₙ be the speed of the airplane relative to the wind.
The distance (d) = 135 miles, Δt = 1 hour, vₐₙ = 135 miles / 1 hour = 135 mph
vₐ = vₙ + vₐₙ
vₐ = vₐₙ
Therefore, vₐ, vₐₙ, vₙ can be represented by an isosceles triangle since vₐ = vₐₙ.
The direction of the wind θ is:
sin(θ / 2) = vₙ / 2vₐ
sin(θ / 2) = 70/ (2*135)
sin(θ / 2) = 0.2593
θ / 2 = sin⁻¹(0.2593) = 15
θ = 30⁰
2α = 180° - 30°
2α = 150°
α = 75°
a) The direction of the wind is 75° in the south east direction while the airplane is heading 30° in the north east direction.