According to the intuitive guess you just made, is the overall momentum of the two-cart system after the collision greater than,
1 answer:
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Refer to the diagram shown below.
For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂ (1)
For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392 (2)
Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N
Answer:
T₂ = 339.06 N
T₃ = 276.57 N
For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂ (1)
For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392 (2)
Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N
Answer:
T₂ = 339.06 N
T₃ = 276.57 N
The sun's intensity for an outer planet located at a distance 6r from the sun is 5.55 W/m². The result is obtained by using the inverse square law formula.
<h3>What is the Inverse Square Law formula?</h3>The Inverse Square Law formula describes the intensity of light is inversely proportional to the square of the distance. It can be expressed as
Where
- I₁ = Intensity at distance 1 (W/m²)
- I₂ = Intensity at distance 2 (W/m²)
- d₁ = distance 1 from a light source (m)
- d₂ = distance 2 from a light source (m)
Given the case the sun's intensity is 200 W/m² for an inner planet at the distance r. If an outer planet is at a distance 6r, what is the sun's intensity?
By using the inverse square law formula, the sun's intensity for an outer planet is
I₂ = 5.55 W/m²
Hence, the sun's intensity for a planet at a distance 6r from the sun is 5.55 W/m².
Learn more about intensity of light here:
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