Answer:
(a) The magnitude of force is 116.6 lb, as exerted by the rod CD
(b) The reaction at A is (-72.7j-38.1k) lb and at B it is (37.5j) lb.
Explanation:
Step by step working is shown in the images attached herewith.
For this given system, the coordinates are the following:
A(0, 0, 0)
B(26, 0, 0)
And the value of angle alpha is 20.95°
Hope that answers the question, have a great day!
Answer:FALSE
Explanation: A negative pressure respirator is a respiratory system which is known to have a low air pressure inside the mask when compared to the air pressure on the outside during Inhalation.
Most of the personal protective equipment (PPE) which are in use in various industries are examples of Negative pressure respirator device,any leak or damage done to the device will allow the inflows of harmful and toxic Air into the person's respiratory system. AIR SUPPLY SYSTEMS ARE KNOWN TO SUPPLY FRESH UNCONTAMINATED AIR THROUGH AIR STORED INSIDE COMPRESSED CYLINDERS OR OTHER SOURCES AVAILABLE.
Answer:
The entropy change of the air is 
Explanation:

is unknown
we can apply the following expression to find 


now substitute

To find entropy change of the air we can apply the ideal gas relationship
Δ

Δ

Δ
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
Answer:
The steady-state temperature difference is 2.42 K
Explanation:
Rate of heat transfer = kA∆T/t
Rate of heat transfer = 6 W
k is the heat transfer coefficient = 152 W/m.K
A is the area of the square silicon = width^2 = (7/1000)^2 = 4.9×10^-5 m^2
t is the thickness of the silicon = 3 mm = 3/1000 = 0.003 m
6 = 152×4.9×10^-5×∆T/0.003
∆T = 6×0.003/152×4.9×10^-5 = 2.42 K