Answer:
Explanation:
From the given information:
Pressure at the elevation 4.8 m = 34.7 kPa
Pressure at the elevation 2.2 m = 57.5 kPa
Recall that:
At the pressure of depth "h";
p = ρgh
where;
ρgh = 34.7 × 10 ³
ρ × 9.81 × h = 34.7 × 10³
ρh = 34.7 × 10³/9.81
ρh = 3.54 × 10³ ----- equation (1)
ρg(h + 4.8 - 2.2) = 57.5 × 10³
ρ × 9.81 × (h + 2.6) = 57.5 × 10³
ρ (h + 2.6) = 57.5 × 10³ / 9.81
ρ (h + 2.6) = 5.86 × 10³ ------ equation (2)
From equation (1) and (2);
The density of the oil ρ is determined to be = 0.8939 × 10³ kg/m³
≅ 893.9 kg/m³
The specific weight w = ρg
w = 0.8939 × 10³ × 9.81
w = 8769.159
w = 8.75 × 10³ kg/m²/s²
The specific gravity SG = [tex]\dfrac{\rho _{oil}}{\rho_{water}}[/tex]
= [tex]\dfrac{0.8939 \times 10^3}{10^3}[/tex]
= 0.8939
≅ 0.894
What is the difference between digital instruments and decimal scaled instruments to measure
Answer Digital measuring instruments are self-contained devices that automatically present the value of the measured quantity on a digital display. And Decimal Scaled Instruments: Record all digits that you can certainly determine from the scale markings and estimate one more digit. I hope this Helped I´m new to this.
Explanation:
After cutting a PVC pipe you should use a
to debure the pipe
Answer:
Deburring Tool
Explanation:
A deburring tool is used in order to debur the PVC pipes. They are mostly used for the plastic pipes.
After the PVC pipes are cut, there are burrs on the pipe surface. To remove these burrs, a deburring tool is used. It removes the burrs form the edges of the PVC pipes that results from grinding, cutting, milling, drilling, etc.
The deburring tools are made from high speed steels.
1: A baseball is hit 4 feet above the ground leaves the bat with an initial speed of 98 ft/sec at an angle of 0 45 is caught by an outfielder at a height of 3 feet.
Answer:
299.36 feet
Explanation:
[tex]To \ find \ the \ distance \ of \ the \ ball \ from \ the \ home \ plate. \\ \\ From \ the \ given \ information:[/tex]
[tex]Height \ h = 4 \ ft[/tex]
[tex]Initial \ speed \ V_o = 98 \ ft/s ec[/tex]
[tex]The \ angle \ \theta = 45^0[/tex]
[tex]Acceleration \ due \ to \ gravity (g)= 32.2 \ ft/s[/tex]
[tex]U_x = V_o \ cos 45 = \dfrac{98}{\sqrt{2}}[/tex]
[tex]U_y = V_o \ sin 45 = \dfrac{98}{\sqrt{2}}[/tex]
So;
[tex]S_y = u_y t - \dfrac{1}{2}gt^2[/tex]
[tex]-1 =\dfrac{98}{\sqrt{2}}t - \dfrac{1}{2}*32*1.85t^2[/tex]
By solving:
[tex]t_1 = 4.32 \ sec[/tex]
Thus;
[tex]horizontal \ distance = U_x t[/tex]
[tex]= \dfrac{98}{\sqrt{2}}\times 4.32[/tex]
[tex]\mathbf{=299.36 \ feet}[/tex]
[tex]\mathbf{Thus \ , the \ distance \ from \ the \ home \ plate \ = \ 299.36 \ feet}[/tex]
A sample of soil has a volume of 0.45 ft^3 and a weight of 53.3 lb. After being dried inan oven, it has a weight of 45.1 lb. It has a specific gravity of solids of 2.70. Compute its moisture content and degree of saturation before it was placed in the oven.
Answer:
a) the moisture content before it was placed in the oven is 18.18%
b) degree of saturation for soil is 72.19%
Explanation:
Given the data in the question;
Moisture Content = [(Weight of soil before dry - dry weight) / dry weight] × 100
so we substitute
Moisture content = [(53.3 - 45.1) / 45.1 ] × 100
= (8.2/45.1) × 100
= 18.18%
Therefore the moisture content before it was placed in the oven is 18.18%
Dry Unit Weight = dry weight / volume
Dry Unit Weight = 45.1 lb / 0.45 ft³
Dry Unit Weight = 100.22 lb/ft³
we know that;
dry unit weight = (Specific gravity × unit weight of water) / (1 + e)
we also know that; unit weight of water is 62.43 lbf/ft³
so we substitute
e = (2.70×62.43 / 100.22) - 1
e = 1.68 - 1
e = 0.68
so void ratio e = 0.68
Now we determine the degree of saturation using the equation;
degree of saturation = (Moisture content × specific gravity) / void ratio
we substitute
degree of saturation = ( 18.18% × 2.7) / 0.68
= 0.49086 / 0.68
= 0.7219 ≈ 72.19%
Therefore degree of saturation for soil is 72.19%
Joe, a technician, is attempting to connect two hubs to add a new segment to his local network. He uses one of his CAT5 patch cables to connect them; however, he is unable to reach the new network segment from his workstation. He can only connect to it from a workstation within that segment. Which of the following is MOST likely the problem?
A. One of the hubs is defective.
B. The new hub is powered down.
C. The patch cable needs to be a CAT6 patch cable.
D. The technician used a straight-through cable.
Answer:
Option D. is correct
Explanation:
Joe uses one of his CAT5 patch cables to connect two hubs to add a new segment to his local network. As he can only connect to it from a workstation within that segment, he is not able to reach the new network segment from his workstation.
The most problem is that the technician used a straight-through cable.
Option D. is correct.
In the United States, a bicyclist is killed:
A. (Every 12 hours
B. Every week
c. Every day
D. Every 6 hours
In the United States, it should be noted that a bicyclist is killed every six hours.
The cause of the accidents has been attributed to the rough driving of vehicle drivers and some faults are on the part of the cyclist as well.
Rapidly overtaking a bicycle is dangerous. Also, there are some vehicle drivers who drive into the lanes that are meant for cyclists. This isn't appropriate.
Drivers should ensure that they are not close to the cyclists when driving as there should be a space of at least 3 feet between the bicycle and the vehicle.
Furthermore, when there is a narrow traffic lane, the vehicle drivers should ensure that there's a clear traffic in the opposite lane before they change their lanes.
Lastly, both the cyclists and the vehicle drivers should not overspeed and drive safely.
Based in the information given above, the correct option is D.
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