At a point on the streamline, Bernoulli's equation is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = density of air, 0.075 lb/ft³ (standard conditions)
g = 32 ft/s²
Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s
Point 2 (stagnation):
At the stagnation point, the velocity is zero.
The density remains constant.
Let p₂ = pressure at the stagnation point.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
= 314.37 lb/ft²
= 314.37/144 = 2.18 lb/in²
Answer: 2.2 psi
Answer:
The maximum energy stored in the combination is 0.0466Joules
Explanation:
The question is incomplete. Here is the complete question.
Three capacitors C1-11.7 μF, C2 21.0 μF, and C3 = 28.8 μF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.
Energy stored in a capacitor is expressed as E = 1/2CtV² where
Ct is the total effective capacitance
V is the supply voltage
Since the capacitors are connected in series.
1/Ct = 1/C1+1/C2+1/C3
Given C1 = 11.7 μF, C2 = 21.0 μF, and C3 = 28.8 μF
1/Ct = 1/11.7 + 1/21.0 + 1/28.8
1/Ct = 0.0855+0.0476+0.0347
1/Ct = 0.1678
Ct = 1/0.1678
Ct = 5.96μF
Ct = 5.96×10^-6F
Since V = 125V
E = 1/2(5.96×10^-6)(125)²
E = 0.0466Joules
Answer:
74.52s
Explanation:
The solution is shown in the picture below
As the question is about changing in frequency of a wave for an observer who is moving relative to the wave source, the concept that should come to our minds is "
Doppler's effect."
Now the general formula of the Doppler's effect is:

-- (A)
Note: We do not need to worry about the signs, as everything is moving towards each other. If something/somebody were moving away, we would have the negative sign. However, in this problem it is not the issue.
Where,
g = Speed of sound = 340m/s.

= Velocity of the receiver/observer relative to the medium = ?.

= Velocity of the source with respect to medium = 0 m/s.

= Frequency emitted from source = 400 Hz.

= Observed frequency = 408Hz.
Plug-in the above values in the equation (A), you would get:


Solving above would give you,

= 6.8 m/s
The correct answer = 6.8m/s
Answer: 14.52*10^6 m/s
Explanation: In order to explain this problem we have to consider the energy conservation for the electron within the coaxial cylidrical wire.
the change in potential energy for the electron; e*ΔV is equal to energy kinetic gained for the electron so:
e*ΔV=1/2*m*v^2 v^=(2*e*ΔV/m)^1/2= (2*1.6*10^-19*600/9.1*10^-31)^1/2=14.52 *10^6 m/s