What scientist improved apon the telescope
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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°.
Answer:
The energy stored in the solenoid is 7.078 x 10⁻⁵ J
Explanation:
Given;
diameter of the solenoid, d = 2.80 cm
radius of the solenoid, r = d/2 = 1.4 cm
length of the solenoid, L = 14 cm = 0.14 m
number of turns, N = 200 turns
current in the solenoid, I = 0.8 A
The cross sectional area of the solenoid is given as;
The inductance of the solenoid is given by;
The energy stored in the solenoid is given by;
E = ¹/₂LI²
E = ¹/₂(2.212 x 10⁻⁴)(0.8)²
E = 7.078 x 10⁻⁵ J
Therefore, the energy stored in the solenoid is 7.078 x 10⁻⁵ J