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belka [17]
3 years ago
12

As an intern with an engineering firm, you are asked to measure the moment of inertia of a large wheel, for rotation about an ax

is through its center. Since you were a good physics student, you know what to do. You measure the diameter of the wheel to be 0.88 m and find that it weighs 280 N . You mount the wheel, using frictionless bearings, on a horizontal axis through the wheel's center. You wrap a light rope around the wheel and hang a 6.32 kg mass from the free end of the rope. You release the mass from rest; the mass descends and the wheel turns as the rope unwinds. You find that the mass has speed 4.0 m/s after it has descended 2.5 m .(a) What is the moment of inertia of the wheel for an axis perpendicular to the wheel at its center?
Physics
1 answer:
AysviL [449]3 years ago
3 0

Answer:

I=2.766\ kg.m^2

Explanation:

We have:

diameter of the wheel, d=0.88\ m

weight of the wheel, w_w=280\ N

mass of hanging object to the wheel, m_o=6.32\ kg

speed of the hanging mass after the descend, v_o=4\ m.s^{-1}

height of descend, h=2.5\ m

(a)

moment of inertia of wheel about its central axis:

I=\frac{1}{2} m.r^2

I=\frac{1}{2} \frac{w_w}{g}.r^2

I=\frac{1}{2} \times \frac{280}{9.8}\times 0.44^2

I=2.766\ kg.m^2

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How many Mars-sized planets can fit in Jupiter
Maru [420]

Answer:

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3 years ago
A straight segment of wire has a length of 25 cm and carries a current of 5A. If the wire is perpendicular to the magnetic field
Irina-Kira [14]

Answer:

The magnitude of the magnetic force acting on the conductor is 0.75 Newton

Explanation:

The parameters given in the question are;

The length of the straight segment of wire, L = 25 cm = 0.25 m

The current carried in the wire, I = 5 A

The orientation of the wire with the magnetic field = Perpendicular

The strength of the magnetic field in which the wire is located, B = 0.60 T

The magnetic force, 'F', is given by the following formula;

F = \underset{I}{\rightarrow }·L×\underset{B}{\rightarrow } = I·L·B·sin(θ)

Where;

\underset{I}{\rightarrow } = The current flowing, I

L = The length of the wire

\underset{B}{\rightarrow } = The magnetic field strength, B

θ = The angle of inclination of the conductor to the magnetic field

Where I = 5 A, L = 0.25 m, B = 0.60 T, and θ = 90°, we get;

F = 5 A × 0.25 m × 0.60 T × sin(90°) = 0.75 N

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The magnitude of the magnetic force, F = 0.75 N.

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