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Westkost [7]
3 years ago
8

A solid disk rotates in the horizontal plane at an angular velocity of 0.649 rpm with respect to an axis perpendicular to the di

sk at its center. The moment of inertia of the disk is 0.101 kg m2. From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance of 0.415 m from the axis. The sand in the ring has a mass of 0.499 kg. After all the sand is in place, what is the angular velocity of the disk
Physics
1 answer:
77julia77 [94]3 years ago
5 0

Answer:

The angular velocity of the disk is 0.0369 rad/sec.

Explanation:

Convert 0.649 rpm to rad/s

0.649 rpm = 0.649 * (2π/60) = 0.0679rad/sec

From the question,

Summation of final angular momentum equals summation of initial angular momentum

I*w = Io*wo

w = wo (lo/I)

But I = Isand + Io

and Isand = Msand * R²sand

Therefore, w = wo (lo/I) = wo (Io/Msand * R²sand + Io)

Where,

wo = 0.0679rad/sec

Io = 0.101 kg m2

Msand = 0.499 kg

Rsand = 0.415 m

Hence,

w = 0.0679 {0.101/(0.499*(0.415)²+0.101)}

w = 0.0679 (0.101/0.1869)

w = 0.0679*0.544 = 0.0369 rad/sec

The angular velocity of the disk is 0.0369 rad/sec.

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Tasya [4]

Answer:

Given the area A of a flat surface and the magnetic flux through the surface \Phi it is possible to calculate the magnitude \frac{\Phi}{A}=B\ cos \theta.

Explanation:

The magnetic flux gives an idea of how many magnetic field lines are passing through a surface. The SI unit of the magnetic flux \Phi is the weber (Wb), of the magnetic field B is the tesla (T) and of the area A is (m^{2}). So 1 Wb=1 T.m².

For a flat surface S of area A in a uniform magnetic field B, with \theta being the angle between the vector normal to the surface S and the direction of the magnetic field B, we define the magnetic flux through the surface as:

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We are told the values of \Phi and B, then we can calculate the magnitude

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3 0
3 years ago
A 1000-kilogram car traveling with a velocity of 20. meters per second decelerates uniformly at -5.0 meters per second2 until it
Elis [28]

Answer:

-20000 kgm/s

Explanation:

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An electric field of 1.32 kV/m and a magnetic field of 0.516 T act on a moving electron to produce no net force. If the fields a
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Answer:

The speed of the electron is 2.55\times 10^3\ m/s.

Explanation:

Given that,

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The magnitude of magnetic field, B = 0.516 T

Both the magnetic and electric fields are acting on the moving electron. Then,  the magnitude of electric field and magnetic field is balanced such that :

evB=eE\\\\v=\dfrac{E}{B}\\\\v=\dfrac{1.32\times 10^3}{0.516}\\\\v=2558.13\ m/s

or

v=2.55\times 10^3\ m/s

So, the speed of the electron is 2.55\times 10^3\ m/s. Hence, this is the required solution.

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