To solve this problem we will apply the first law of thermodynamics which details the relationship of energy conservation and the states that the system's energy has. Energy can be transformed but cannot be created or destroyed.
Accordingly, the rate of work done in one cycle and the heat transferred can be expressed under the function,
Substitute 1W for 
and 1.5 W for 
Now calculcate the rate of specific internal energy increase,
The rate of specific internal energy increase is 1.6667W/kg
If she has a choice and the wiring details are stated on the packaging,
then Janelle should look for lights that are wired in parallel within the
string, and she should avoid lights that are wired in series within the string.
If a single light in a parallel string fails, then only that one goes out.
The rest of the lights in the string continue to shimmer and glimmer.
If a single light in a series string fails, then ALL of the lights in that string
go out, and it's a substantial engineering challenge to determine which light
actually failed.
Answer;
B. Increased levels of carbon dioxide, a greenhouse gas, leads to increased phytoplankton growth.
Explanation;
-A combination of warm water, high nutrient levels, and adequate sunlight may cause a harmful algae bloom. These blooms may damage aquatic ecosystems by blocking sunlight and depleting oxygen that other organisms need to survive.
-Algae blooms have been increasing globally, and climate change may be playing a role in the increment. For instance, during the warm summer season or when water is warmer, some harmful types of algae to grow faster than other, more benign varieties.
-Additionally, the warmer surface water also prevents water from mixing vertically, allowing algae to grow thicker and faster.
The work done on the barbell is -165.62 Nm.
Explanation:
Work done on any object is the measure of force required to move that object from one position to another. So it is determined by the product of force acting on the object with the displacement of the object.
In the present problem, the displacement of the object on acting of force is given as 1.3 m. And the weight of the object which is a barbel is given as 13 kg. As the work is to lift the object from the ground, so the acceleration due to gravity will be acting on the object. In other words, the force applied on the object to lift it should be in opposite direction to the acting of acceleration due to gravity.
Thus, 
Now, the force is -127.4 N and the displacement is 1.3 m.
So, 
So, the work done on the barbell is -165.62 Nm.
The answer would be 21.6 but rounded up it would be 22J.
