Answer:
Mechanical Advantage Formula
The efficiency of a machine is equal to the ratio of its output to its input. It is also equal to the ratio of the actual and theoretical MAs. But, it does not mean that low-efficiency machines are of limited use. An automobile jack, for example, have to overcome a great deal of friction and therefore it has low efficiency. But still, it is extremely valuable because small effort can be applied to lift a great weight.
Also, in another way the mechanical advantage is the force generated by a machine to the force applied to it which is applied in assessing the performance of the machine.
The mechanical advantage formula is:
MA = FBFA
Explanation:
MAmechanical advantageFBthe force of the object
FAthe effort to overcome the force
Answer:
Tech A
Explanation:
The amount of energy required to apply the same force with a 1:1 ratio is divided into 4, so you can apply 4 times as much force than a 1:1 ratio. efficiency and speed come into play here, but assuming the machine powering the gear can run at a unlimited RPM, 4:1 will have more force and a slower output speed than a 2:1 ratio.
Answer:
The graph representing the linear inequalities is attached below.
Explanation:
The inequalities given are :
y>x-2 and y<x+1
For tables for values of x and y and get coordinates to plot for both equation.
In the first equation;
y>x-2
y=x-2
y-x = -2
The table will be :
x y
-2 -4
-1 -3
0 -2
1 -1
2 0
The coordinates to plot are : (-2,-4) , (-1,-3), (0,-2), (1,-1) ,(2,0)
Use a dotted line and shade the part right hand side of the line.
Do the same for the second inequality equation and plot then shade the part satisfying the inequality.
The graph attached shows results.
Answer:
a. Rotational speed of the drill = 375.96 rev/min
b. Feed rate = 75 mm/min
c. Approach allowance = 3.815 mm
d. Cutting time = 0.67 minutes
e. Metal removal rate after the drill bit reaches full diameter. = 9525 mm³/min
Explanation:
Here we have
a. N = v/(πD) = 15/(0.0127·π) = 375.96 rev/min
b. Feed rate = fr = Nf = 375.96 × 0.2 = 75 mm/min
c. Approach allowance = tan 118/2 = (12.7/2)/tan 118/2 = 3.815 mm
d. Approach allowance T∞ =L/fr = 50/75 = 0.67 minutes
e. R = 0.25πD²fr = 9525 mm³/min.
