Acceleration of gravity on jupiter
2 answers:
You might be interested in
Complete Question:
A hollow cylinder with an inner radius of 4.0mm and an outer radius of 30mm conducts a 3.0-A current flowing parallel to the axis of the cylinder. If the current density is uniform throughout the wire, what is the magnitude of the magnetic field at a point 12mm from its center?
Answer:
The magnitude of the magnetic field = 7.24 μT
Explanation:
Inner radius, a = 4.0 mm = 0.004 m
Outer radius, b = 30 mm = 0.03 m
Radius, r = 12 mm = 0.012 m
let h² = b² - a²
h² = 0.03² - 0.004²
h² = 0.000884
Let d² = r² - a²
d² = 0.012² - 0.004²
d² = 0.000128
Current I = 3A
μ = 4π * 10⁻⁷
The magnitude of the magnetic field is given by:
B = 7.24 * 10⁻⁶T
B = 7.24 μT
Explanation:
Given that,
Mass of the ball, m = 1.2 kg
Initial speed of the ball, u = 10 m/s
Height of the floor from ground, h = 32 m
(a) Let v is the final speed of the ball. It can be calculated using the conservation of energy as :
v = -25.04 m/s (negative as it rebounds)
The impulse acting on the ball is equal to the change in momentum. It can be calculated as :
J = -42.048 kg-m/s
(b) Time of contact, t = 0.02 s
Let F is the average force on the floor from by the ball. Impulse acting on an object is given by :
F = 0.8409 N
Hence, this is the required solution.