A ball is thrown vertically upward from the ground. Its distance in feet from the ground in t seconds is s equals negative 16 t
1 answer:
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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
Answer:
θ = 19.66°
Explanation:
To determine the angle that the rope makes with the vertical for the two people, you first take into account the potential energy of the first person before he swings on the rope:
h: distance to the ground
g: gravitational acceleration = 9.8m/s^2
m: mass of the first person = 55 kg
In the image attache below you can notice that the height h is:
Then, the potential energy is:
When the first person picks up the second person (when the rope is exactly vertical), all the potential energy becomes kinetic energy. Next, when both people reaches the maximum height h' the energy must be equal to the initial potential energy of the first person:
From the previous equation you can get h':
Finally, you obtain the angle between the rope at the height h,' and the vertical, by calculating the following:
hence, the angle between the rope and the vertical, when the two people are in the rope is 19.66°
