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]
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.
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%