Answer:
D
Explanation:
Because I just had that answer
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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Answer:
1) The greatest height attained by the ball equals 20.387 meters.
2) The time it takes for the ball to reach 15 meters approximately equals 1 second.
Explanation:
The greatest height will be attained when the ball stop's in the air and starts falling back to the earth.
thus using third equation of kinematics we obtain the height attained as

where
'v' is the final speed of the ball
'u' is the initial speed of the ball
'a' is the acceleration that the ball is under which in this case equals 9.81 
's' is the distance it covers
Thus for maximum height applying the values in the equation we get

Using the same equation we can find the speed of the ball when it reaches 15 meters of height as

the time it takes to reduce the velocity to this value can be found by first equation of kinematics as

Answer:
Option E is correct 310N
Explanation:
Given that the force used to push the crate is F = 200N
The force directed 20° below the horizontal
Mass of crate is m = 25kg
Weight of the crate can be determine using
W = mg
g is gravitational constant =9.8m/s²
W = 25×9.8
W = 245 N
Check attachment. For free body diagram and better understanding
Using newton second law along the vertical axis since we want to find the normal force
ΣFy = m•ay
ay = 0, since the body is not moving in the vertical or y direction
N—W—F•Sin20 = 0
N = W+F•Sin20
N = 245+ 200Sin20
N = 245 + 68.4
N = 313.4 N
The normal force is approximately 310 N to the nearest ten