Answer:
0.5°c
Explanation:
Humidity ratio by mass can be expressed as
the ratio between the actual mass of water vapor present in moist air - to the mass of the dry air
Humidity ratio is normally expressed in kilograms (or pounds) of water vapor per kilogram (or pound) of dry air.
Humidity ratio expressed by mass:
x = mw / ma (1)
where
x = humidity ratio (kgwater/kgdry_air, lbwater/lbdry_air)
mw = mass of water vapor (kg, lb)
ma = mass of dry air (kg, lb)
It can be as:
x = 0.005 (100) / [(100 - 100)]
x = 0.005 x 100 / (100 - 100)
x = 0.005 x 100 / 0
x = 0.5°c
So the temperature to which atmospheric air must be cooled in order to have humidity ratio of 0.005 lb/lb is 0.5°c
Answer:
False
Explanation:
The government decides the productions.
Answer:
No, they need to be somewhat flexible so that forces such as turbulance don't shear the wing off.
Answer: a. Leave the lane closest to the emergency as soon as it is safe to do so, or slow down to a speed of 20 MPH below the posted speed limit.
Explanation:
Giving a way to the law enforcement vehicle and a medical emergency vehicle is necessary. If one approaches an emergency vehicle parked along the roadway one should change the lane as the vehicle may not move and the driver may also waste his or her time also one should also slow down his or her speed while approaching the vehicle as most of the emergency vehicle are in rush to reach the hospital so the driver should maintain some distance with the medical emergency vehicle.
Answer:
The flexural strength of a specimen is = 78.3 M pa
Explanation:
Given data
Height = depth = 5 mm
Width = 10 mm
Length L = 45 mm
Load = 290 N
The flexural strength of a specimen is given by
![\sigma = \frac{3 F L}{2 bd^{2} }](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Cfrac%7B3%20F%20L%7D%7B2%20bd%5E%7B2%7D%20%7D)
![\sigma = \frac{3(290)(45)}{2 (10)(5)^{2} }](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Cfrac%7B3%28290%29%2845%29%7D%7B2%20%2810%29%285%29%5E%7B2%7D%20%7D)
78.3 M pa
Therefore the flexural strength of a specimen is = 78.3 M pa