I did some fire simulation work recently where a small paper fire (think like half a ream of printer paper) was putting out 4000 watts, or about 5 horsepower. A large roaring campfire is going to be a few multiples of that depending on size, of course.
Answer:
The picture on the right with more circles inside of it is more dense
Explanation:
Answer:
C. attaching an active metal to make the pipe the cathode in an electrochemical cell.
Explanation:
Cathodic protection is a technique which helps in controlling the increased rate of corrosion of a metal surface by making it the cathode of an electrochemical cell. It connects the metal to be protected to a more easily corroded which is usually referred to as the sacrificial metal to act as the anode.
This technique preserves the metal by providing a highly active metal that can act as an anode and provide free electrons. By introducing these free electrons, the active metal sacrifices its ions and keeps the less active steel from corroding.
Answer:
5.75 L.
Explanation:
From the question given above, the following data were obtained:
Rate = 54.1 miles/gallons
Distance = 132 km
Volume (in L) consumed =?
Next, we shall convert 132 km to mile. This can be obtained as follow:
1 km = 0.621 mile
Therefore,
132 km = 132 km × 0.621 mile / 1 km
132 km = 81.972 mile
Next, we shall determine the volume (in gallons) of the gas needed. This can be obtained as follow:
Rate = 54.1 miles/gallons
Distance = 81.972 mile
Volume (in gallon) =?
Rate = Distance / volume
54.1 = 81.972 / volume
Cross multiply
54.1 × volume = 81.972
Divide both side by 54.1
Volume = 81.972 / 54.1
Volume = 1.52 gallon.
Finally, we shall convert 1.52 gallon to litre (L). This can be obtained as follow:
1 gallon = 3.785 L
Therefore,
1.52 gallon = 1.52 gallon × 3.785 L / 1 gallon
1.52 gallon = 5.75 L
Therefore, 5.75 L of the gas will be consumed.