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
It us 200 energy if it the right awnser
Answer:
y = 33.93 10⁵ m
Explanation:
This is an interference exercise, for the contributory interference is described by the expression
d sin θ = m λ
let's use trigonometry for the angle
tan θ = y / L
how the angles are small
tan θ = sin θ / cos tea = sin θ
we substitute
sin θ = y / L
d y / L = m λ
y = m λ L / d
the light fulfills the relation of the waves
c = λ f
λ = c / f
λ = 3 10⁸ /375
λ = 8 10⁵ m
first order m = 1
let's calculate
y = 1 8 10⁵ 4030 10-9 / 950 10-9
y = 33.93 10⁵ m
Answer:
PE=mgh
M= Mass (kg)
G= Gravitational field strength (N/kg)
H= Hight (m)
PE= Gravitational Potential Energy (J)
Explanation:
Gravitational Potential Energy is the energy stored in a object due to its position above the Earth's surface.
Answer:
C. Technician B
Explanation:
Excessive Galvanic activity:
To check for excessive galvanic activity, voltmeter is used to check the coolant. If the voltmeter is giving a reading greater than 0.5 V, there is excessive galvanic activity. Excessive galvanic activity is solved by flushing the coolant fluid from engine and refiling it.
Electrolysis problem:
When the system is not properly ground, the cooling system accepts stray current and the coolant becomes an electrolyte which might eat up the radiator. To test for excessive electrolysis, start the engine and turn on all electrical accessories, if the reading is more than 0.5 V, there is electrolysis problem. Ground wires and connections should be checked at this point to stop stray current.
In our case, the first reading is 0.2 V(engine turned off) which is normal and there is no excessive galvanic activity. This means that Technician A is not correct. The second reading is 0.8 V when the engine and all electrical accessories are turned on. This reading is greater than 0.5 V which means there is an electrolysis problem. This means that Technician B is correct and ground wires and connections should be inspected and repaired.
