2. Physics - Q&A

PART66Online: Good addition, thanks. 2n^2 gives 2 for the first shell and 8 for the second. I have added the formula to the question explanation for clarity.

PART66Online: Hello. Indeed, moved to module 2.

claus: 15 psi + 14.7 (atmospheric pressure)=29.7 almost 30

Ricky HO: OK, thank you

Eddz25: ideal gas law and general gas law is just the same. you will use the these formulas depending on what is given

Eddz25: s=ut + 1/2 at(2)

karol288: Ek=0.5*m*v^2 Ek=0.5*2*12^2 Ek=0,5*2*144=144 J

Angel: *Liquid matter is characterized by molecules that have more energy and increased movement. This is what the book says.

Angel: Joules per kilogram per Kelvin(Jkg^-1 K^-1).

Angel: The confusion arises from the difference between calculating work done by the machine versus calculating the efficiency of the machine. Your physics book's content focuses heavily on efficiency, and one of the ways to calculate it is indeed: Efficiency=(Mechanical Advantage (MA)/Speed Ratio)*100 This formula links mechanical advantage to the input/output relationship and how well the machine minimizes losses. The book also emphasizes that no machine is 100% efficient, as losses always occur due to heat, friction, etc. The Test Question Your test question was specifically about work done by a machine, assuming it is 100% efficient. For this, the mechanical advantage and input are often used because: Mechanical Advantage relates the force output to the force input: Mechanical Advantage (MA)=Output Force/Input Force or Mechanical Advantage (MA)= Input Force/Output Force To calculate work input, you need: Work Input=Input Force*Distance Input. When the machine is 100% efficient: Work Output=Work Input. Work done by the machine can then be derived using mechanical advantage (how the forces are transformed) and the input work provided. Thus, "Mechanical Advantage and Input" is correct if you're focused on determining the relationship between the forces and work in a theoretical, ideal (100% efficient) scenario. Why the Two Answers Differ The physics book emphasizes real-world efficiency (with inevitable losses), while the test assumes an idealized scenario (no losses). This difference changes the approach: For calculating efficiency: You use both output and input parameters (e.g., measured performance and ideal performance). For determining work in a 100% efficient machine: You relate input (force/distance) and mechanical advantage directly, which is why the test prioritizes "Mechanical Advantage and Input." The key lies in understanding the context of the question: The test assumes an idealized situation (100% efficiency). The book explains real-world machines, where efficiency is less than 100%.

disho251: thank you

mano: Weight arm momentum 10kg 4m 40 5kg 2m 10 Total aft 50 then 20kg 2.5m 50

piss monger: Perch l'indice di rifrazione dato dalla velocit della luce nel vuoto fratto la velocit della luce nel mezzo, in questo caso sai che nel vuoto la luce va a 1,5 volte la velocit che avrebbe nella fibra. Per trovare la velocit nel mezzo ti baster fare (v luce nel vuoto / indice di rifrazione)

Jordi: thank you man i managed to pass all the modules with the help of this website.

Trevian: 1 kW is approximately equal to: 3413 BTU per hour 1.341 HP (horsepower) 360,000 joules

Trevian: i don't understand this question.

Diverdeep: should be written in another way, in this way, I understand 4.5 is the nominal power of the motor, not the effective work (output)

fgibbons2018: Power = Work / Time

George: youtu.be/LMSTNYe4A1g

Jordi: thanks man

Matar: the formula is T(F)= (t( degrees C)*9/5)+32