if a student lifts their weight of 450 newtons up a set of stairs 5 meters high, how much work did they do?
2 answers:
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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
To solve this problem it is necessary to apply the concepts related to the frequency in a spring, the conservation of energy and the total mechanical energy in the body (kinetic or potential as the case may be)
PART A) By definition the frequency in a spring is given by the equation
Where,
m = mass
k = spring constant
Our values are,
k=1700N/m
m=5.3 kg
Replacing,
PART B) To solve this section it is necessary to apply the concepts related to the conservation of energy both potential (simple harmonic) and kinetic in the spring.
Where,
k = Spring constant
m = mass
y = Vertical compression
v = Velocity
This expression is equivalent to,
Our values are given as,
k=1700 N/m
V=1.70 m/s
y=0.045m
m=5.3 kg
Replacing we have,
Solving for A,
PART C) Finally, the total mechanical energy is given by the equation