They are experiencing the pull of leaving the atmosphere
Given:
density of air at inlet, 
density of air at inlet, 
Solution:
Now,
(1)
where
A = Area of cross section
= velocity of air at inlet
= velocity of air at outlet
Now, using eqn (1), we get:
= 1.14
% increase in velocity = 
=114%
which is 14% more
Therefore % increase in velocity is 14%
Answer:
when u find out pls lmk! i have the same question and I've been stuck for a while lol
Answer:
25.71 kgm/s
Explanation:
Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.
Given that K₂ = 0.7K₁
1/2mv₂² = 0.7(1/2mv₁²)
v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s
Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.
Th magnitude of object A's momentum change is thus 25.71 kgm/s
Answer:
* The first thing we observe is that the frequency response does not change
* The current that circulates in the circuit decreases due to the new resistance at the resonance point,
Z = R + R₂
Explanation:
The impedance of a series circuit is
Z₀² = R² + (X_L-X_C) ²
when we place another resistor in series the initial resistance impedance changes to
Z² = (R + R₂) ² + (X_L - X_C) ²
let's analyze this expression
* The first thing we observe is that the frequency response does not change
* The current that circulates in the circuit decreases due to the new resistance at the resonance point,
Z = R + R₂
