Answer:
The correct option is;
c. Leaving the chuck key in the drill chuck
Explanation:
A Common safety issues with a drill press leaving the chuck key in the drill chuck
It is required that, before turning the drill press power on, ensure that chuck key is removed from the chuck. A self ejecting chuck key reduces the likelihood of the chuck key being accidentally left in the chuck.
It is also required to ensure that the switch is in the OFF position before turning plugging in the power cable
Be sure that the chuck key is removed from the chuck before turning on the power. Using a self-ejecting chuck key is a good way of insuring that the key is not left in the chuck accidentally. Also to avoid accidental starting, make sure the switch is in the OFF position before plugging in the cord. Always disconnect the drill from the power source when making repairs.
Answer:
The pressure drop is 269.7N/m^2
Explanation:
∆P = ∆h × rho × g
∆h = 3.2cm = 3.2/100 = 0.032m, rho = 860kg/m^3, g = 9.8m/s^2
∆P = 0.032×860×9.8 = 269.7N/m^2
Answer:
16 seconds
Explanation:
Given:
C = 60
L = 4 seconds each = 4*4 =16
In this problem, the first 3 timing stages are given as:
200, 187, and 210 veh/h.
We are to find the estimated effective green time of the fourth timing stage. The formula for the estimated effective green time is:
Let's first find the fourth stage critical lane group ratio
, using the formula:


Solving for
, we have:
Let's also calculate the volume capacity ratio X,

X = 0.704
For the the estimated effective green time of the fourth timing stage, we have:
Substituting figures in the equation, we now have:
15.78 ≈ 16 seconds
The estimated effective green time of the fourth timing stage is 16 seconds
Answer:
P = 18035.25 N
Explanation:
Given
D = 10.4 mm
ΔD = 3.2 ×10⁻³ mm
E = 207 GPa
ν = 0.30
If
σ = P/A
A = 0.25*π*D²
σ = E*εx
ν = - εz / εx
εz = ΔD / D
We can get εx as follows
εz = ΔD / D = 3.2 ×10⁻³ mm / 10.4 mm = 3.0769*10⁻⁴
Now we find εx
ν = - εz / εx ⇒ εx = - εz / ν = - 3.0769*10⁻⁴ / 0.30 = - 1.0256*10⁻³
then
σ = E*εz = (207 GPa)*(-1.0256*10⁻³) = - 2.123*10⁸ Pa
we have to obtain A:
A = 0.25*π*D² = 0.25*π*(10.4*10⁻3)² = 8.49*10⁻⁵ m²
Finally we apply the following equation in order o get P
σ = P/A ⇒ P = σ*A = (- 2.123*10⁸Pa)*(8.49*10⁻⁵ m²) = 18035.25 N