Answer: 2.92 s
Explanation:
Given
Mass of ball is
The initial velocity of the ball is
Velocity after the rebound is
Force during the contact is
We know, change in momentum is Impulse
Thus, the force is applied for 2.92 s
A straight wire segment 5 m long makes an angle of 30° with a uniform magnetic field of 0.37 T. Find the magnitude of the force
1 answer:
Answer:
The magnitude of the force on the wire is 2.68 N.
Explanation:
Given that,
Length of the wire, L = 5 m
Magnetic field, B = 0.37 T
Angle between wire and the magnetic field,
Current in the wire, I = 2.9 A
We need to find the magnitude of the force on the wire. The magnetic force in the wire is given by :
So, the magnitude of the force on the wire is 2.68 N. Hence, this is the required solution.
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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:0.13ml
Explanation:
