Answer:
Explanation:
Given that,
Surface area A= 17m²
The speed at the top v" = 66m/s
Speed beneath is v' =40 m/s
The density of air p =1.29kg/m³
Weight of plane?
Assuming that,
the height difference between the top and bottom of the wind is negligible and we can ignore any change in gravitational potential energy of the fluid.
Using Bernoulli equation
P'+ ½pv'²+ pgh' = P'' + ½pv''² + pgh''
Where
P' is pressure at the bottom in N/m²
P" is pressure at the top in N/m²
v' is velocity at the bottom in m/s
v" is velocity at the top in m/s
Then, Bernoulli equation becomes
P'+ ½pv'² = P'' + ½pv''²
Rearranging
P' — P'' = ½pv"² —½pv'²
P'—P" = ½p ( v"² —v'²)
P'—P" = ½ × 1.29 × (66²-40²)
P'—P" = 1777.62 N/m²
Lift force can be found from
Pressure = force/Area
Force = ∆P ×A
Force = (P' —P")×A
Since we already have (P'—P")
Then, F=W = (P' —P")×A
W = 1777.62 × 17
W = 30,219.54 N
The weight of the plane is 30.22 KN
Answer:
Explanation:
Given the height reached by a balloon after t sec modeled by the equation
h=1/2t²+1/2t
a) To calculate the height of the balloon after 40 secs we will substitute t = 40 into the modeled equation and calculate the value of t
If h(t)=1/2t²+1/2t
h(40) = 1/2(40)²+1/2 (40)
h(40) = 1600/2 + 40/2
h(40) = 800 + 20
h(40) = 820 feet
The height of the balloon after 40 secs is 820 feet
b) Velocity is the change of displacement of a body with respect to time.
v = dh/dt
v(t) = 2(1/2)t²⁻¹ + 1/2
v(t) = t + 1/2
when v = 0sec
v(0) = 0 + 1/2
v(0) = 1/2 ft/sec
at v = 30secs
v(30) = 30 + 1/2
v(30) = 30 1/2 ft/sec
average velocity = v(30) - v(0)
average velocity = 30 1/2 - 1/2
average velocity of the balloon between t = 0 and t = 30 = 30 ft/sec
c) Velocity is the change of displacement of a body with respect to time.
v = dh/dt
v(t) = 2(1/2)t²⁻¹ + 1/2
v(t) = t + 1/2
The velocity of the balloon after 30secs will be;
v(30) = 30+1/2
v(30) = 30.5ft/sec
The velocity of the balloon after 30 secs is 30.5 feet/sec
Answer: 363 Ω.
Explanation:
In a series AC circuit excited by a sinusoidal voltage source, the magnitude of the impedance is found to be as follows:
Z = √((R^2 )+〖(XL-XC)〗^2) (1)
In order to find the values for the inductive and capacitive reactances, as they depend on the frequency, we need first to find the voltage source frequency.
We are told that it has been set to 5.6 times the resonance frequency.
At resonance, the inductive and capacitive reactances are equal each other in magnitude, so from this relationship, we can find out the resonance frequency fo as follows:
fo = 1/2π√LC = 286 Hz
So, we find f to be as follows:
f = 1,600 Hz
Replacing in the value of XL and Xc in (1), we can find the magnitude of the impedance Z at this frequency, as follows:
Z = 363 Ω
