Answer:
1.3cm
Explanation:
the arrow is 3 lines past the 1 so it is 1.3cm
Answer:
the maximum length of the specimen before deformation is 0.4366 m
Explanation:
Given the data in the question;
Elastic modulus E = 124 GPa = 124 × 10⁹ Nm⁻²
cross-sectional diameter D = 4.2 mm = 4.2 × 10⁻³ m
tensile load F = 1810 N
maximum allowable elongation Δl = 0.46 mm = 0.46 × 10⁻³ m
Now to calculate the maximum length
for the deformation, we use the following relation;
= [ Δl × E × π × D² ] / 4F
so we substitute our values into the formula
= [ (0.46 × 10⁻³) × (124 × 10⁹) × π × (4.2 × 10⁻³)² ] / ( 4 × 1810 )
= 3161.025289 / 7240
= 0.4366 m
Therefore, the maximum length of the specimen before deformation is 0.4366 m
Answer:
DIAMETER = 9.797 m
POWER = 
Explanation:
Given data:
circular windmill diamter D1 = 8m
v1 = 12 m/s
wind speed = 8 m/s
we know that specific volume is given as

where v is specific volume of air
considering air pressure is 100 kPa and temperature 20 degree celcius

v = 0.8409 m^3/ kg
from continuity equation





mass flow rate is given as


the power produced ![\dot W = \dot m \frac{ V_1^2 - V_2^2}{2} = 717.3009 [\frac{12^2 - 8^2}{2} \times \frac{1 kJ/kg}{1000 m^2/s^2}]](https://tex.z-dn.net/?f=%5Cdot%20W%20%3D%20%5Cdot%20m%20%5Cfrac%7B%20V_1%5E2%20-%20V_2%5E2%7D%7B2%7D%20%3D%20717.3009%20%5B%5Cfrac%7B12%5E2%20-%208%5E2%7D%7B2%7D%20%5Ctimes%20%5Cfrac%7B1%20kJ%2Fkg%7D%7B1000%20m%5E2%2Fs%5E2%7D%5D)

Answer:
I know ship has stream line body
Answer:
as slated in your solution, if delay time is 2.30 mins, hence 9 vehicle will be on queue as the improved service commenced.
Explanation:
4 vehicle per min, in 2 mins of the delay time 8 vehicles while in 0.3 min average of 1 vehicle join the queue. making 9 vehicle maximum