Answer:
Efficiency = 52%
Explanation:
Given:
First stage
heat absorbed, Q₁ at temperature T₁ = 500 K
Heat released, Q₂ at temperature T₂ = 430 K
and the work done is W₁
Second stage
Heat released, Q₂ at temperature T₂ = 430 K
Heat released, Q₃ at temperature T₃ = 240 K
and the work done is W₂
Total work done, W = W₁ + W₂
Now,
The efficiency is given as:

or
Work done = change in heat
thus,
W₁ = Q₁ - Q₂
W₂ = Q₂ - Q₃
Thus,

or

or

also,

or

thus,

thus,

or

or
Efficiency = 52%
Answer:
a) 2.5 m/s²
b) 6.12 m/s
Explanation:
Tension of rope = T = 356N
Weight of material = W = 478 N
Distance from the ground = s = 7.5 m
Acceleration due to gravity = g = 9.81 m/s²
Mass of material = m = 478/9.81 = 48.72
Final velocity before the bundle hits the ground = v
Initial velocity = u = 0
Acceleration experienced by the material when being lowered = a
a) W-T = ma
⇒478-356 = 48.72×a

⇒a = 2.5 m/s²
∴ Acceleration achieved by the material is 2.5 m/s²
b) v²-u² = 2as
⇒v²-0 = 2×2.5×7.5
⇒v² = 37.5
⇒v = 6.12 m/s
∴ Velocity of the material before hitting the ground is 6.12 m/s
Explanation:
Kepler’s third law states that for all objects orbiting a given body, the cube of the semimajor axis (A) is proportional to the square of the orbital period (P).
For each of our planets orbiting the Sun, the relationship between the orbital period and semimajor axis can be represented by the equation as:

k is constant of proportionality
It is required to solve the above equation for k

Answer:
E = 1.655 x 10⁷ N/C towards the filament
Explanation:
Electric field due to a line charge is given by the expression
E =
[/tex]
where λ is linear charge density of line charge , r is distance of given point from line charge and ε₀ is a constant called permittivity and whose value is
8.85 x 10⁻¹².
Putting the given values in the equation given above
E = 
E = 1.655 x 10⁷ N/C