Answer:
In terms of distance, average speed is 20 km/h
In terms of displacement, average speed is 0 km/h.
Explanation:
Total distance:
Total time is 3.0 hours
but:
In terms of distance.
substitute:
In terms of displacement:
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The diagram is missing; however, we know that the intensity of a sound wave is inversely proportional to the square of the distance from the source:
where I is the intensity and r is the distance from the source.
We can assume for instance that the initial distance from the source is r=1 m, so that we put
The intensity at r=3 m will be
Therefore, the sound intensity has decreased by a factor
.
where I is the intensity and r is the distance from the source.
We can assume for instance that the initial distance from the source is r=1 m, so that we put
The intensity at r=3 m will be
Therefore, the sound intensity has decreased by a factor
Answer:
(a) The package lands 682 meters horizontally ahead from the point the package was dropped from the plane
(b) The horizontal component = 39.0 m/s
The vertical component = 171.55 m/s
(c) The angle of impact is 77.19°
Explanation:
The parameters given are;
Velocity of the plane, vₓ = 39.0 m/s
Height of the plane above the ground, h = 1.50 × 10² m = 1,500 m
(a) The time, t, before the package hits the ground when dropped from the plane is given by the relation;
h = u·t + 1/2×g×t²
Where:
g = Acceleration due to gravity = 9.81 m/s²
u = Initial vertical velocity = 0 m/s
Hence;
1500 = 0×t + 1/2 × 9.81 × t² = 4.905·t²
∴ t = √(1500/4.905) = 17.49 s
The horizontal distance the package travels before landing = 17.49 × 39 ≈ 682 m
The package lands 682 meters horizontally ahead from the point the package was dropped from the plane
(b) The vertical velocity, , of the package just before landing is given by the relation;
² = u² + 2·g·h
u = 0 m/s
∴ ² = 0 + 2×9.81×1500 = 29430 m²/s²
= √29430 = 171.55 m/s
Hence the horizontal component = 39.0 m/s
The vertical component = 171.55 m/s
(c) The angle of impact, θ, is given as follows;
∴ θ = tan⁻¹(4.4) = 77.19°.