Answer:
Time constant of RC circuit is 0.105 seconds.
Explanation:
It is given that,
Resistance, 
Capacitance, 
We need to find the expected time constant for this RC circuit. It can be calculated as :
So, the time constant for this RC circuit is 0.105 seconds. Hence, this is the required solution.
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
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Answer:
Explanation:
For the cat to stay in place on the merry go round without sliding the magnitude of maximum static friction must be equal to magnitude of centripetal force
Where the r is the radius of merry-go-round and v is the tangential speed
but
So we have
Substitute the given values
So
Answer:
a) -1.25 rev/s² and 23.3 rev
b) 2.67s
Explanation:
a) ω
= (500 rev/min)(1min/ 60s) => 8.333 rev/s
ω
= (200 rev/min)(1min/ 60s) => 3.333rev/s
time 't'= 4 s
angular acceleration 'α'=?
constant angular acceleration equation is given by,
ω= ω + αt
α= (ω - ω )/t => (3.333-8.333)/4
α= -1.25 rev/s²
θ-θ
= ω t + 1/2αt²
=(8.333)(4) + 1/2 (-1.25)(4)²
=23.3 rev
b) ω=0 (comes to rest)
ω = 3.333 rev/s
α= -1.25 rev/s²
ω= ω + αt
t= (ω - ω)/α => (0- 3.333)/-1.25
t= 2.67s
