From Carnot's theorem, for any engine working between these two temperatures:
efficiency <= (1-tc/th) * 100
Given: tc = 300k (from question assuming it is not 5300 as it seems)
For a, th = 900k, efficiency = (1-300/900) = 70%
For b, th = 500k, efficiency = (1-300/500) = 40%
For c, th = 375k, efficiency = (1-300/375) = 20%
Hence in case of a and b, efficiency claimed is lesser than efficiency calculated, which is valid case and in case of c, however efficiency claimed is greater which is invalid.
Answer: Is this a question? I dont understand.
Explanation:
Answer:
Explanation:
= Mg ion = 
= F ion = 
q = Charge of electron = 
r = Distance between ions = 
k = Coulomb constant = 
Electrical force is given by
![F=-\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=-\dfrac{8.99\times 10^9\times 2\times 1.6\times 10^{-19}\times -1\times 1.6\times 10^{-19}}{[(0.072+0.133)\times 10^{-9}]^2}\\\Rightarrow F=1.09527\times 10^{-8}\ N](https://tex.z-dn.net/?f=F%3D-%5Cdfrac%7Bkq_1q_2%7D%7Br%5E2%7D%5C%5C%5CRightarrow%20F%3D-%5Cdfrac%7B8.99%5Ctimes%2010%5E9%5Ctimes%202%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%5Ctimes%20-1%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%7D%7B%5B%280.072%2B0.133%29%5Ctimes%2010%5E%7B-9%7D%5D%5E2%7D%5C%5C%5CRightarrow%20F%3D1.09527%5Ctimes%2010%5E%7B-8%7D%5C%20N)
The attractive force is
Answer:
KE_2 = 3.48J
Explanation:
Conservation of Energy
E_1 = E_2
PE_1+KE_1 = PE_2+KE_2
m*g*h+(1/2)m*v² = m*g*h+(1/2)m*v²
(0.0780kg)*(9.81m/s²)*(5.36m)+(.5)*(0.0780kg)*(4.84m/s)² = (0.0780kg)*(9.81m/s²)*(2m)+KE_2
4.10J+0.914J = 1.53J + KE_2
5.01J = 1.53J + KE_2
KE_2 = 3.48J
- In order to achieve the desired resistance under the given circumstances, we would connect two 50 Ohms resistors in parallel and then connect it in series with the 20 Ohms resistors.
- In order to get a 35 Ohms resistance under the given circumstances, we would connect two 50 Ohms resistors in parallel and then connect it in series with two 20 Ohms resistors that are connected in parallel.
<h3>How to achieve the desired resistance under these circumstances?</h3>
In order to achieve the desired resistance under the given circumstances, we would connect two 50 Ohms resistors in parallel and then connect it in series with the 20 Ohms resistors.
Mathematically, the total equivalence resistance of two resistors that are connected in parallel is given by:
1/Rt = 1/R₁ + 1/R₂
1/Rt = 1/50 + 1/50
1/Rt = 2/50
1/Rt = 1/25
Rt = 25 Ohms.
Next, we would connect this 25 Ohms resistor in series with the 20 Ohms resistor:
R₃ = 20 + Rt
R₃ = 20 + 25
R₃ = 45 Ohms.
<h3>Part B.</h3>
In order to get a 35 Ohms resistance under the given circumstances, we would connect two 50 Ohms resistors in parallel and then connect it in series with two 20 Ohms resistors that are connected in parallel.
1/Rt = 1/R₁ + 1/R₂
1/Rt = 1/50 + 1/50
1/Rt = 2/50
1/Rt = 1/25
Rt = 25 Ohms.
1/R't = 1/R₁ + 1/R₂
1/R't = 1/20 + 1/20
1/R't = 2/20
1/R't = 1/10
R't = 10 Ohms.
Next, we would connect the 25 Ohms resistor in series with the 10 Ohms resistor:
R₃ = 10 + Rt
R₃ = 10 + 25
R₃ = 35 Ohms.
Read more on resistors in parallel here: brainly.com/question/15121871
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Complete Question:
You need a 45-ω resistor, but the stockroom has only 20-ω and 50-ω resistors.
(a) How can the desired resistance be achieved under these circumstances?
(b) What can you do if you need a 35-ω resistor?
