Answer:
![\Delta S=1.69J/K](https://tex.z-dn.net/?f=%5CDelta%20S%3D1.69J%2FK)
Explanation:
We know,
..............(1)
where,
η = Efficiency of the engine
T₁ = Initial Temperature
T₂ = Final Temperature
Q₁ = Heat available initially
Q₂ = Heat after reaching the temperature T₂
Given:
η =0.280
T₁ = 3.50×10² °C = 350°C = 350+273 = 623K
Q₁ = 3.78 × 10³ J
Substituting the values in the equation (1) we get
![0.28=1-\frac{Q_2}{3.78\times 10^{3}}](https://tex.z-dn.net/?f=0.28%3D1-%5Cfrac%7BQ_2%7D%7B3.78%5Ctimes%2010%5E%7B3%7D%7D)
or
![\frac{Q_2}{3.78\times 10^3}=0.72](https://tex.z-dn.net/?f=%5Cfrac%7BQ_2%7D%7B3.78%5Ctimes%2010%5E3%7D%3D0.72)
or
![Q_2=3.78\times 10^3\times0.72](https://tex.z-dn.net/?f=Q_2%3D3.78%5Ctimes%2010%5E3%5Ctimes0.72)
⇒ ![Q_2 =2.721\times 10^3 J](https://tex.z-dn.net/?f=Q_2%20%3D2.721%5Ctimes%2010%5E3%20J)
Now,
The entropy change (
) is given as:
![\Delta S=\frac{\Delta Q}{T_1}](https://tex.z-dn.net/?f=%5CDelta%20S%3D%5Cfrac%7B%5CDelta%20Q%7D%7BT_1%7D)
or
![\Delta S=\frac{Q_1-Q_2}{T_1}](https://tex.z-dn.net/?f=%5CDelta%20S%3D%5Cfrac%7BQ_1-Q_2%7D%7BT_1%7D)
substituting the values in the above equation we get
![\Delta S=\frac{3.78\times 10^{3}-2.721\times 10^3 J}{623K}](https://tex.z-dn.net/?f=%5CDelta%20S%3D%5Cfrac%7B3.78%5Ctimes%2010%5E%7B3%7D-2.721%5Ctimes%2010%5E3%20J%7D%7B623K%7D)
![\Delta S=1.69J/K](https://tex.z-dn.net/?f=%5CDelta%20S%3D1.69J%2FK)
Answer:
D) F
Explanation:
Let m and M be the mass of the balls A and B respectively and r be the distance between the two balls. The magnitude of attractive gravitational force experienced by the balls due to each other is given by the relation :
......(1)
Now, if the masses of both the balls gets doubled as well as there separation distance also gets doubled, then let F₁ be the new gravitational force acting on them.
Since, New mass of ball A = 2M
New mass of ball b = 2m
Distance between the two balls = 2r
Substitute these values in equation (1).
![F_{1} =\frac{G(2M)(2m)}{(2r)^{2} }](https://tex.z-dn.net/?f=F_%7B1%7D%20%3D%5Cfrac%7BG%282M%29%282m%29%7D%7B%282r%29%5E%7B2%7D%20%7D)
![F_{1} =\frac{4GMm}{4r^{2} }=\frac{GMm}{r^{2} }](https://tex.z-dn.net/?f=F_%7B1%7D%20%3D%5Cfrac%7B4GMm%7D%7B4r%5E%7B2%7D%20%7D%3D%5Cfrac%7BGMm%7D%7Br%5E%7B2%7D%20%7D)
Using equation (1) in the above equation.
F₁ = F
1. b An element is the simplest form of a substance.
3. a Atoms are comprised of particles such as protons, neutrons, and electrons.
2. c Atoms are tiny pieces of matter.
4. d Phase refers to the state of matter: gas, liquid, and solid.
Answer:
25%
Explanation:
If there is 20% scrap, then
80% of (production + scrap) = 72 pcs/hr
Now,
![\frac{80}{100} \times (72+scrap)= 72 pcs/hr\\0.80\times (72+Scrap) = 72 \\ 72+Scrap = \frac{72}{0.80}\\72 + Scrap = 90\\Scrap = 18 pcs /hr](https://tex.z-dn.net/?f=%5Cfrac%7B80%7D%7B100%7D%20%5Ctimes%20%2872%2Bscrap%29%3D%2072%20pcs%2Fhr%5C%5C0.80%5Ctimes%20%2872%2BScrap%29%20%3D%2072%20%5C%5C%2072%2BScrap%20%3D%20%5Cfrac%7B72%7D%7B0.80%7D%5C%5C72%20%2B%20Scrap%20%3D%2090%5C%5CScrap%20%3D%2018%20pcs%20%2Fhr)
The percent increase in labor productivity is,
.
Or it can be written as 25% of which would give 80 pieces per hour.