Answer:
267.07 km
Explanation:
We have given the radius of the earth = 6378.1 km
In 2014 the difference between the magnetic north pole and geographical north pole is 2.40°
2.40°
We know that linear distance is given by
So we have to travel 267.07 km in going from magnetic north pole to geographic north pole
Answer:
2649600 Joules
Explanation:
Efficiency = 40%
m = Mass of air = 92000 kg
v = Velocity of wind = 12 m/s
Kinetic energy is given by
The kinetic energy of the wind is 6624000 Joules
The wind turbine extracts 40% of the kinetic energy of the wind
The energy extracted by the turbine every second is 2649600 Joules
Answer:
See explanation
Explanation:
Solution:-
- Here we will assume that the grating has the line density ( N ) defined by the number of lines per mm.
- The angle that each fringe forms on the screen is defined by ( θ ).
- The order of bright/dark spot is defined by an integer ( n )
- The wavelength of the incident light is ( λ )
- Here we will use the relation given for diffraction grating by the Young's Experiment as follows:
- The above given formulation is for constructive interference.
- We will inspect the effect of increasing the distance between the screen and the grating. Consider the length ( L ) from the center of the grating to the center of the screen. The distance ( yn ) will denote the distance between each fringe in vertical direction on the screen.
- For small angles ( θ ) we can make an approximation of sin ( θ ) ≈ tan ( θ ). Where,
sin ( θ ) ≈ tan ( θ ) = [ yn / L ]
- Substitute the above approximation in the given relation of diffraction gratings as follows:
- To double the distance between the screen and grating we will use the above relation with ( 2L ):
yn ∝ L
Result: The distance between each order of bright and dark fringe is doubled. The interference pattern would have twice the spread! This also means that less number of bright spots would be seen on the screen as the coverage area would require a larger screen to accommodate the entire interference pattern. The spread also reduces the intensity/contrast between the bright and dark fringes because the distance travelled by each ray of light has increased. The intensity is inversely proportional to the square of distance travelled.
- Similarly, the line density of the grating ( N ) was doubled. Then,
yn ∝ N
Result: The distance between each order of bright and dark fringe is doubled. The interference pattern would have twice the spread!This also means that less number of bright spots would be seen on the screen as the coverage area would require a larger screen to accommodate the entire interference pattern.
The time per lap was calculated by measuring the time for seven laps and dividing the total time by seven.
total time
It is given that the precision is of 0.1 s. it means it is correct upto 1 place beyond decimal.
So, the actual value could vary from 457.800 s to 457.899 s i.e. the time per lap could be 65.400 s to 65.414 s
It means measured time 65.414 s has maximum error of 0.021%. Hence, the measured value is quite precise.
Answer:
(a) The gauge pressure is 4,885.3 Pa
(b) The anchoring force needed to hold the elbow in place is approximately 296.5 N
The direction of the anchoring force is approximately ≈ 134.8
Explanation:
The given parameters in the fluid dynamics question are;
Pipe elbow angle, θ = 90°
The mass flowrate of the water, = 40 kg/s
The diameter of the elbow, D = 10 cm
The pressure at the point of discharge of the fluid = Atmospheric pressure
The elevation between the inlet and exit of the elbow, z = 50 cm
The weight of the elbow and the water = Negligible
(a) The velocity of the fluid, v₁ = v₂ = v is given as follows;
ρ·v·A =
v = /(ρ·A)
Where;
ρ = The density of the fluid (water) = 997 kg/m³
A = The cross-sectional area of the of the elbow = π·D²/4 = π×0.1²/4 ≈ 0.00785398163
A = 0.00785398163 m²
v = 40 kg/s/(0.00785398163 m² × 997 kg/m³) ≈ 5.1083 m/s
Bernoulli's equation for the flow of fluid is presented as follows;
v₁ = v₂ for the elbow of uniform cross section
P₁ - P₂ = ρ·g·(z₂ - z₁)
P₁ - P₂ = The gauge pressure =
z₂ - z₁ = z = 50 cm = 0.5 m
∴ = 997 kg/m³ × 9.8 m/s² × 0.5 m = 4,885.3 Pa
The gauge pressure, = 4,885.3 Pa
(b) The forces acting on the elbow are;
+
·A = -β·
·v
= β·
·v
∴ = -β·
·v -
·A
= -1.03 × 40 kg/s × 5.1083 m/s - 4,885.3 Pa × 0.00785398163 m² ≈ -208.83 N
= 1.03 × 40 kg/s × 5.1083 m/s = 210.46196 N
The resultant force, , is given as follows;
= √(
² +
²)
∴ = √((-208.83)² + (210.46196)²) ≈ 296.486433934
Therefore;
The anchoring force needed to hold the elbow in place, ≈ 296.5 N
The direction of the anchoring force, θ
∴ θ = arctan(210.46196/(-208.83)) = -45..2230044°
θ = 180° + -45.2230044° = 134.7769956°
∴ The direction of the anchoring force is approximately, θ ≈ 134.8°