Answer:
a. 2x/3
b. 8
Explanation:
fundamental period can be defined to mean that at after every period of 2π radians or 360° the value of graph is repeated. For such functions the fundamental period is the period after which they repeat themselves.
It van also be looked as The fundamental period of cos(θ) is 2π. That is (for example) cos(0) to cos(2π) represents one full period.
Please see attachment for the step by step solution.
Answer:
M2 = 0.06404
P2 = 2.273
T2 = 5806.45°R
Explanation:
Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.
Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,
To1 = (1.008)*(1000) = 1008 ºR
R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)
F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga
For the air q = cp(To2– To1)
(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2
Table A.3 of steam table gives P/P* = 2.273,
T/T* = 0.2066,
To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =
F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit
The question is asking whether that statement is true or false. Options are;
A) True
B) False
This is about usage of Swing arm restraints.
<em><u>B) False</u></em>
There are different safety features that people employ when a vehicle is lifted. However, for this question, we will only talk about swing arm restraints.
- Swing arm restraints are lifting restraint devices that are used to prevent a cars arms from shifting or going out of position after that car has been lifted and mounted.
- This swing arm restraint does not prevent a vehicle from falling off a lift as it just helps to ensure that the swing arms that are unloaded basically maintain their position.
Read more at; brainly.com/question/17972874
Answer:
hello your question is incomplete attached below is the complete question
answer: There is a hydraulic jump
Explanation:
First we have to calculate the depth of flow downstream of the gate
y1 = 
----------- ( 1 )
Cc ( concentration coefficient ) = 0.61 ( assumed )
Yg ( depth of gate opening ) = 0.5
hence equation 1 becomes
y1 = 0.61 * 0.5 = 0.305 m
calculate the flow per unit width q
q = Q / b ----------- ( 2 )
Q = 10 m^3 /s
b = 2 m
hence equation 2 becomes
q = 10 / 2 = 5 m^2/s
next calculate the depth before hydraulic jump y2 by using the hydraulic equation
answer : where y1 < y2 hence a hydraulic jump occurs in the lined channel
attached below is the remaining part of the solution
