Answer:
the claim is not valid or reasonable.
Explanation:
In order to test the claim we will find the maximum and actual efficiencies. maximum efficiency of a heat engine can be found as:
η(max) = 1 - T₁/T₂
where,
η(max) = maximum efficiency = ?
T₁ = Sink Temperature = 300 K
T₂ = Source Temperature = 400 K
Therefore,
η(max) = 1 - 300 K/400 K
η(max) = 0.25 = 25%
Now, we calculate the actual frequency of the engine:
η = W/Q
where,
W = Net Work = 250 KJ
Q = Heat Received = 750 KJ
Therefore,
η = 250 KJ/750 KJ
η = 0.333 = 33.3 %
η > η(max)
The actual efficiency of a heat engine can never be greater than its Carnot efficiency or the maximum efficiency.
<u>Therefore, the claim is not valid or reasonable.</u>
Answer:
<em>Maximum=70 m</em>
<em>Minimum=26 m</em>
Explanation:
<u>Vector Addition
</u>
Since vectors have magnitude and direction, adding them takes into consideration not only the magnitudes but also their respective directions. Two vectors can be totally collaborative, i.e., point to the same direction, or be totally opposite. In the first case, the magnitude of the sum is at maximum. Otherwise, it's at a minimum.
Thus, the maximum magnitude of the sum is 48+22 = 70 m and the minimum magnitude of the sum is 48-22= 26 m
To solve this problem we will use the concepts related to power, defined as the amount of energy applied over a period of time.
The energy in this case is the accumulated in the form of potential energy, over a period of time. Thus we will have that the mathematical expression of the power can be expressed as
Here,
E = Energy
t = time
As the energy is equal to the potential Energy we have tat
The weight (mg) of the man is 700N, the height (h) is 8m and the time is 10s, then:
Therefore the correct answer is A.
<span>Assuming continuous operation (24/7), we can say that
Energy produced : Energy per hour * 24 (number of hours in a day) - 365 (number of days in a year.
Energy per hour: 2050 * 1.055 = 2162.75 kg.
So, we proceed to calculate the results
E: 2162.75 * 24 * 365 = 18,945,690 kj per year.
Now, we transform kj to megajoule, remembering that kilo is 10*3 and mega is 1'*6, so we divide the result by 1,000 in order to get the results in megajoules, and the answer would be:
18,945.69 megajoules can be produced per year.</span>
