Answer:
a) The magnitude of the thrust provided by the jet's engines is 4840 newtons.
b) The magnitude of the tension in the cable connecting the jet and glider is 572 newtons.
Explanation:
a) By Newton's laws we construct the following equations of equilibrium. Please notice that both the glider and the jet experiments has the same acceleration:
Jet
(1)
Glider
(2)
Where:
- Thrust of jet engines, measured in newtons.
- Tension in the cable connecting the jet and glider, measured in newtons.
, 
- Masses of the glider and the jet, measured in kilograms.
- Acceleration of the glider-jet system, measured in meters per square second.
If we know that 
, 
and 
, then the solution of this system of equations:
By (2):
By (1):
The magnitude of the thrust provided by the jet's engines is 4840 newtons.
b) The magnitude of the tension in the cable connecting the jet and glider is 572 newtons.
Move with constant speed or accelerate and will determine direction
When air is blown into the open pipe,
L = 
where nis any integral number 1,2,3,4 etc. and λ is the wavelength of the oscillation
⇒λ=
Note here that n=1 is for fundamental, n=2 is first harmonic and so on..
⇒ third harmonic will be n=4
Given L=6m, n=4, solving for λ we get:
λ=
=3m
Relationship of frequency(f), velocity of sound (c) and wavelength(λ) is:
c=f.λ Or f= 
⇒f=
≈115 Hz
Answer:
It will be more than deta t
Explanation:
Because
deta t' = န deta t
But န= 1/√ (1 - v²/c²
So the observers in all the initial frames will be more than deta t
