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prohojiy [21]
2 years ago
5

when approaching the front of an idling jet engine, the hazard area extends forward of the engine approximately

Physics
1 answer:
Dominik [7]2 years ago
7 0

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.

Learn more about turbine engine refer

brainly.com/question/807662

#SPJ4

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What kind of surface is suitable for infared radiation
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8 0
3 years ago
Determine the magnitude of the resultant force acting on a 1.5 −kg particle at the instant t=2 s, if the particle is moving alon
Phoenix [80]

Answer:

F = 63N

Explanation:

M= 1.5kg , t= 2s, r = (2t + 10)m and

Θ = (1.5t² - 6t).

magnitude of the resultant force acting on 1.5kg = ?

Force acting on the mass =

∑Fr =MAr

Fr = m(∇r² - rθ²) ..........equation (i)

∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)

The horizontal path is defined as

r = (2t + 10)

dr/dt = 2, d²r/dt² = 0

Angle Θ is defined by

θ = (1.5t² - 6t)

dθ/dt = 3t, d²θ/dt² = 3

at t = 2

r = (2t + 10) = (2*(2) +10) = 14

but dr/dt = 2m/s and d²r/dt² = 0m/s

θ = (1.5(2)² - 6(2) ) = -6rads

dθ/dt =3(2) - 6 = 0rads

d²θ/dt = 3rad/s²

substituting equation i into equation ii,

Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)

∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]

∑F = 1.5(14*3+0) = 63N

F = √(Fr² +FΘ²) = √(0² + 63²) = 63N

7 0
3 years ago
Which characteristic of a sound is affected by the amount of energy used to create that sound? In what direction the sound will
olya-2409 [2.1K]

Answer:

The correct option is;

How loud or soft the sound is

Explanation:

The loudness of a sound wave is given by the amount of energy that the pressure wave carries and it is measured in decibels (dB) which is the relative intensity of the pressure wave of a sound to the standard pressure

A loud sound has a high amplitude and a soft sound has a low amplitude, such that as the amplitude of the sound is increased, due to increased energy input, the sound becomes louder, and as the amplitude of the sound is decreased due to reduced energy input, the sound becomes softer.

5 0
3 years ago
The sound produced by the loudspeaker in the drawing has a frequency of 11999 Hz and arrives at the microphone via two different
const2013 [10]

The speed at which sound travels through the gas in the tube is 719.94m/s

<u>Explanation:</u>

Given:

Frequency, f = 11999Hz

Wavelength, λ = 0.03m

Velocity, v = ?

Sound speed in the tube is calculated by multiplying the frequency v by the wavelength λ.

As the sound loudness changed from a maximum to a minimum, then we know the sound interference in the case changed from constructive interference (the two sound waves are in phase, i.e. peaks are in a line with peaks and so the troughs), to a destructive interference (peaks coinciding with troughs). The least distance change required to cause such a change is a half wavelength distance, so:

λ/2 = 0.03/2

 λ  = 0.06m

We know,

v = λf

v = 0.06 X 11999Hz

v = 719.94m/s

Therefore, the speed at which sound travels through the gas in the tube is 719.94m/s

3 0
3 years ago
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