Answer:
A) ≥ 325Kpa
B) ( 265 < Pe < 325 ) Kpa
C) (94 < Pe < 265 )Kpa
D) Pe < 94 Kpa
Explanation:
Given data :
A large Tank : Pressures are at 400kPa and 450 K
Throat area = 4cm^2 , exit area = 5cm^2
<u>a) Determine the range of back pressures that the flow will be entirely subsonic</u>
The range of flow of back pressures that will make the flow entirely subsonic
will be ≥ 325Kpa
attached below is the detailed solution
<u>B) Have a shock wave</u>
The range of back pressures for there to be shock wave inside the nozzle
= ( 265 < Pe < 325 ) Kpa
attached below is a detailed solution
C) Have oblique shocks outside the exit
= (94 < Pe < 265 )Kpa
D) Have supersonic expansion waves outside the exit
= Pe < 94 Kpa
A)We know the formula of the angular speed is ω = 2π / TWhere T is the time period.When second hand completes one revolution then the time taken is 60s.So T = 60sThen the angular speed of the second hand is ω= 2π / (60s) = 0.1047 rad/sb)When the minute hand completes one revolution the time taken is T = 1 hr = 3600sThen the angular speed of the minute hand is ω =(2π) / (3600s) = 0.001745 rad/sc)When the hour hand completes one revolution then the timeperiod is T = 12hrs = (12)(3600)sThen the angular speed of the hour hand is ω =(2π) / [(12)(3600)s] = 1.45444 x 10^-4 rad/s
The kilogram is the Standard International System of Units unit of mass. It is defined as the mass of a particular international prototype made of platinum-iridium and kept at the International Bureau of Weights and Measures.
Increased by a factor of 4
Answer:
Explanation:
distance of shuttle from centre of the earth = radius of the orbit
= 6300 + 300 = 6600 km
= 6600 x 10³
Formula of time period of the satellite
T = 2π R /v₀ , v₀ is orbital velocity
v₀ = √gR , ( if height is small with respect to radius )
T = 2π R /√gR
= 2π√ R /√g
= 2 x 3.14 x √ 6600 x 10³ / √9.8
= 2 x 3.14 x 256.9 x 10 / 3.13
= 5154.41 s
= 5154.41 / 60 minutes
= 85.91 m
85.9 minutes.
2 ) No of sunrise per day = no of rotation per day
= 24 x 60 / 85.9
= 16.76
or 17 sunrises.
