The proposed grading at a project site will consist of 25,100 m3 of cut and 23,300 m3 of fill and will be a balanced earthwork j
1 answer:
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Answer:
Explanation:
Given that :
The initial volume of water = 1000.00 mL = 1000000 mm³
The initial temperature of the water = 10° C
The height of the water column h = 1000.00 mm
The final temperature of the water = 70° C
The coefficient of thermal expansion for the glass is ∝ =
The objective is to determine the the depth of the water column
In order to do that we will need to determine the volume of the water.
We obtain the data for physical properties of water at standard sea level atmospheric from pressure tables; So:
At temperature the density of the water is
At temperature the density of the water is
The mass of the water is
Thus; we can say ;
⇒
Thus, the volume of the water after heating to a required temperature of is 1022400 mm³
However; taking an integral look at this process; the volume of the water before heating can be deduced by the relation:
The area of the water before heating is:
The area of the heated water is :
Finally, the depth of the heated hot water is:
Hence the depth of the heated hot water is
Answer:
Speed of aircraft ; (V_1) = 83.9 m/s
Explanation:
The height at which aircraft is flying = 3000 m
The differential pressure = 3200 N/m²
From the table i attached, the density of air at 3000 m altitude is; ρ = 0.909 kg/m3
Now, we will solve this question under the assumption that the air flow is steady, incompressible and irrotational with negligible frictional and wind effects.
Thus, let's apply the Bernoulli equation :
P1/ρg + (V_1)²/2g + z1 = P2/ρg + (V_2)²/2g + z2
Now, neglecting head difference due to high altitude i.e ( z1=z2 ) and V2 =0 at stagnation point.
We'll obtain ;
P1/ρg + (V_1)²/2g = P2/ρg
Let's make V_1 the subject;
(V_1)² = 2(P1 - P2)/ρ
(V_1) = √(2(P1 - P2)/ρ)
P1 - P2 is the differential pressure and has a value of 3200 N/m² from the question
Thus,
(V_1) = √(2 x 3200)/0.909)
(V_1) = 83.9 m/s
Answer:
D. N= 11. 22 rad/s (CW)
Explanation:
Given that
Form factor R = 8
Speed of sun gear = 5 rad/s (CW)
Speed of ring gear = 12 rad/s (CW)
Lets take speed of carrier gear is N
From Algebraic method ,the relationship between speed and form factor given as follows
here negative sign means that ring and sun gear rotates in opposite direction
Lets take CW as positive and ACW as negative.
Now by putting the values
N= 11. 22 rad/s (CW)
So the speed of carrier gear is 11.22 rad/s clockwise.