Answer:
Maximum slope for hand-propelled wheelchair ramps should be 1" of rise to every 12" of length (4.8 degree angle; 8.3% grade).
Maximum slope for power chairs should be 1.5" rise to 12" length (7.1 degree angle; 12.5% grade).
Minimum width should be 36" (inside rails) - (48" is ideal).
The "deck" or surface of the ramp should be set down between a side-rail assembly such that there is about a 2" curb or lip along the edges of the ramp surface. Decking could consist of 1" X 6" pressure treated pine, (or 3/4" pressure treated plywood applied to a frame).
If possible, the end of the deck (where it meets the lower ground surface) should be beveled to provide a smooth transition from the ramp to level ground. Alternatively, a sheet of 10 Ga. steel at least 10" long and sized to fit the width of the ramp could be used to span the space between the deck surface and the walk or driveway surface at the end of the ramp. This piece should overlap the ramp deck by 2" and be fastened securely with 4 large countersunk flat-head wood screws.
A level platform of at least 5' X 5' should be at the top of ramp to allow for wheelchair maneuvering. If the entrance way opens outward, there should be 1' of surface area extending from the side of the door opening to allow motion to the side without backing the chair during door opening. This landing should not be considered part of the overall "run"/length of the ramp. Any turning point along the ramp needs a level landing. If the turn is a right angle (90 degrees), the landing should be a minimum of 5' by 4'. If a "switchback" of 180 degrees is constructed, the level landing should measure at least 5' X 8'. Ramps longer than 30' should provide a platform every 30' for purposes of safety and to create opportunity for rest
Step-by-step explanation:
Answer: 1200pi cubic cm per sec
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Work Shown:
dr/dt = 3 is the rate of change of the radius
V = (4/3)pi*r^3 ..... volume of a sphere
dV/dt = (4/3)*3*pi*r^2*dr/dt ... chain rule
dV/dt = 4*pi*r^2*dr/dt
dV/dt = 4*pi*r^2*3
dV/dt = 12*pi*r^2
dV/dt = 12*pi*10^2 ... plug in r = 10
dV/dt = 1200pi
Let’s suppose PT is equal to TQ
Therefore the equations will be equal
3x+7=8x-8
3x-8x=-7-8
-5x=-15
x=3
We have found x now we will substitute it into each equation
PT=3(3)+7
PT=9+7
PT=16
Then we will substitute the second equation
TQ=8(3)-8
TQ=24-8
TQ=16
And here is the answer please correct me if I’m wrong
9514 1404 393
Answer:
34.5 square meters
Step-by-step explanation:
We assume you want to find the area of the shaded region. (The actual question is not visible here.)
The area of the triangle (including the rectangle) is given by the formula ...
A = 1/2bh
The figure shows the base of the triangle is 11 m, and the height is 1+5+3 = 9 m. So, the triangle area is ...
A = (1/2)(11 m)(9 m) = 49.5 m^2
The rectangle area is the product of its length and width:
A = LW
The figure shows the rectangle is 5 m high and 3 m wide, so its area is ...
A = (5 m)(3 m) = 15 m^2
The shaded area is the difference between the triangle area and the rectangle area:
shaded area = 49.5 m^2 - 15 m^2 = 34.5 m^2
The shaded region has an area of 34.5 square meters.
