Answer:
2.27secs
Explanation:
The car will take 2.27secs to reach a speed of 10 ft/s.
Kindly go through the attached files for a step by step solution that will show you how the answer was gotten and the normal reaction of both the front and rear wheels.
Answer:
volumetric flow rate =
Velocity in pipe section 1 =
velocity in pipe section 2 = 12.79 m/s
Explanation:
We can obtain the volume flow rate from the mass flow rate by utilizing the fact that the fluid has the same density when measuring the mass flow rate and the volumetric flow rates.
The density of water is = 997 kg/m³
density = mass/ volume
since we are given the mass, therefore, the volume will be mass/density
25/997 =
volumetric flow rate =
Average velocity calculations:
<em>Pipe section A:</em>
cross-sectional area =
mass flow rate = density X cross-sectional area X velocity
velocity = mass flow rate /(density X cross-sectional area)
<em>Pipe section B:</em>
cross-sectional area =
mass flow rate = density X cross-sectional area X velocity
velocity = mass flow rate /(density X cross-sectional area)
Complete Question:
A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of width 5mm and a depth of cut of 0.5 mm. Estimate the minimum time required to reduce the depth of the plate by 20 mm if the tool moves at 400 mm per second.
Answer:
26 mins 40 secs
Explanation:
Reduction in depth, Δd = 20 mm
Depth of cut,
Number of passes necessary for this reduction,
n = 20/0.5
n = 40 passes
Tool width, w = 5 mm
Width of metal plate, W = 200 mm
For a reduction in the depth per pass, tool will travel W/w = 200/5 = 40 times
Speed of tool, v = 100 mm/s
minimum time required to reduce the depth of the plate by 20 mm:
number of passes * Time/pass
n * Time/pass
40 * 40
1600 = 26 mins 40 secs