Answer:
The camera would flow the same velocity and acceleration of the droplet until 9ft. After that it would switch gear and accelerate back to its original spot at the rate of 140.54 m/s2 in order to catch the next droplet.
Explanation:
The time it takes for the drop to reach 7-ft and 9-ft at the acceleration of 32.2ft/s2:
When s = 7ft, 
When s = 9ft, 
So it would take Δt = 0.75 - 0.66 = 0.09 s for each drop to travel from 7ft to 9ft. As the drops are released every 0.5s, this means that after our cam follow the first drop to the end of the 7-9ft window, it has 0.5 - 0.09 = 0.41 s or less to reverse its acceleration and go back to the original spot to take care of the next drop.
So at 9ft, the drop and camera velocity would be:
So if the camera reverses its direction to go back to original spot at the rate of a (ft/s2) and initial speed of 24.1 ft/s. Within the time span of 0.41s to catch the next drop at 7ft
So the camera would flow the same velocity and acceleration of the droplet until 9ft. After that it would switch gear and accelerate back to its original spot at the rate of 140.54 m/s2 in order to catch the next droplet.
Answer: 0.93 mA
Explanation:
In order to calculate the current passing through the water layer, as we have the potential difference between the ends of the string as a given, assuming that we can apply Ohm’s law, we need to calculate the resistance of the water layer.
We can express the resistance as follows:
R = ρ.L/A
In order to calculate the area A, we can assume that the string is a cylinder with a circular cross-section, so the Area of the water layer can be written as follows:
A= π(r22 – r12) = π( (0.0025)2-(0.002)2 ) m2 = 7.07 . 10-6 m2
Replacing by the values, we get R as follows:
R = 1.4 1010 Ω
Applying Ohm’s Law, and solving for the current I:
I = V/R = 130 106 V / 1.4 1010 Ω = 0.93 mA
Answer:
The correct answer is "25.03 mm".
Explanation:
Given:
Thickness of plate,
= 10 mm
On a side,
= 100 mm
Diameter hole,
= 25 mm
Coefficient of thermal expansion,
CTE = 
Now,
⇒ 
= 
= 
= 
hence,
The final diameter of hole will be:
Answer:
(d) None. No provisions exist.
Explanation:
B&P Code § 6738 prohibits a non-licensed person from being the sole proprietor of an engineering business. The non-licensed can be a partner in an engineering business that offers civil, electrical, or mechanical services. It is mandatory that at least one licensed engineer must be a co-owner of the business.
