Answer:
Δ h = 52.78 m
Explanation:
given,
Atmospheric pressure at the top of building = 97.6 kPa
Atmospheric pressure at the bottom of building = 98.2 kPa
Density of air = 1.16 kg/m³
acceleration due to gravity, g = 9.8 m/s²
height of the building = ?
We know,
Δ P = ρ g Δ h
(98.2-97.6) x 10³ = 1.16 x 9.8 x Δ h
11.368 Δ h = 600
Δ h = 52.78 m
Hence, the height of the building is equal to 52.78 m.
Point A has the largest magnitude of acceleration as compared to other points on the position verses time graph.
On the graph, A is the point where magnitude of the acceleration of the particle is greatest as compared to other positions on the graph because the height of point A is the largest as compared to other points of the graph.
The graph shows at which point acceleration of an object is higher and lower so we can conclude that point A has the largest magnitude of acceleration as compared to other points on the position verses time graph.
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Answer:
Time of flight A is greatest
Explanation:
Let u₁ , u₂, u₃ be their initial velocity and θ₁ , θ₂ and θ₃ be their angle of projection. They all achieve a common highest height of H.
So
H = u₁² sin²θ₁ /2g
H = u₂² sin²θ₂ /2g
H = u₃² sin²θ₃ /2g
On the basis of these equation we can write
u₁ sinθ₁ =u₂ sinθ₂=u₃ sinθ₃
For maximum range we can write
D = u₁² sin2θ₁ /g
1.5 D = u₂² sin2θ₂ / g
2 D =u₃² sin2θ₃ / g
1.5 D / D = u₂² sin2θ₂ /u₁² sin2θ₁
1.5 = u₂ cosθ₂ /u₁ cosθ₁ ( since , u₁ sinθ₁ =u₂ sinθ₂ )
u₂ cosθ₂ >u₁ cosθ₁
u₂ sinθ₂ < u₁ sinθ₁
2u₂ sinθ₂ / g < 2u₁ sinθ₁ /g
Time of flight B < Time of flight A
Similarly we can prove
Time of flight C < Time of flight B
Hence Time of flight A is greatest .
Answer:
detecting and indicating an electric current