when approaching the front of an idling jet engine, the hazard area extends forward of the engine approximately 25 feet.
<h3>What impact, if any, would jet fuel and aviation gasoline have on a turbine engine?</h3>
Tetraethyl lead, which is present in gasoline, deposits itself on the turbine blades. Because jet fuel has a higher viscosity than aviation gasoline, it may retain impurities with greater ease.
Once the gasoline charge has been cleared, start the engine manually or with an electric starter while cutting the ignition and using the maximum throttle.
On the final approach, the aeroplane needs to be re-trimmed to account for the altered aerodynamic forces. A substantial nose-down tendency results from the airflow producing less lift on the wings and less downward force on the horizontal stabiliser due to the reduced power and slower velocity.
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The ball because the Kinetic Energy transfers from the bat to the ball, increasing the movement and acceleration of the ball because of the Kinetic Energy transferred from the origin force (The bat)
Answer:
220 A
Explanation:
The magnetic force on the floating rod due to the rod held close to the ground is F = BI₁L where B = magnetic field due to rod held close the ground = μ₀I₂/2πd where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, I₂ = current in rod close to ground and d = distance between both rods = 11 mm = 0.011 m. Also, I₁ = current in floating rod and L = length of rod = 1.1 m.
So, F = BI₁L
F = (μ₀I₂/2πd)I₁L
F = μ₀I₁I₂L/2πd
Given that the current in the rods are the same, I₁ = I₂ = I
So,
F = μ₀I²L/2πd
Now, the magnetic force on the floating rod equals its weight , W = mg where m = mass of rod = 0.10kg and g = acceleration due to gravity = 9.8 m/s²
So, F = W
μ₀I²L/2πd = mg
making I subject of the formula, we have
I² = 2πdmg/μ₀L
I = √(2πdmg/μ₀L)
substituting the values of the variables into the equation, we have
I = √(2π × 0.011 m × 0.1 kg × 9.8 m/s²/[4π × 10⁻⁷ H/m × 1.1 m])
I = √(0.01078 kgm²/s²/[2 × 10⁻⁷ H/m × 1.1 m])
I = √(0.01078 kgm²/s²/[2.2 × 10⁻⁷ H])
I = √(0.0049 × 10⁷kgm²/s²H)
I = √(0.049 × 10⁶kgm²/s²H)
I = 0.22 × 10³ A
I = 220 A
Only if a force acts upon it, it can move.
Solution:
The relation between the potential difference and the electric field between the plates of the parallel plate capacitor is given by :
Differentiating on both the sides with respect to time, we get
Therefore, the rate of the electric field changes between the plates of the parallel plate capacitor is given by :
V/m-s
