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A uniform disk is constrained to rotate about an axis passing through its center and perpendicular to the plane of the disk. If

katrin2010 [14]

A uniform disk is constrained to rotate about an axis passing through its center and perpendicular to the plane of the disk. If the disk starts from rest and is then brought in contact with a spinning rubber wheel, we observe that the disk gradually begins to rotate too. If after 35 s of contact with this spinning rubber wheel, the disk has an angular velocity of 4.0 rad/s, find the average angular acceleration that the disk experiences. (Assume the positive direction is in the initial direction of rotation of the disk. Indicate the direction with the sign of your answer.)

Assume after 35 s of contact with this spinning rubber wheel, the disk has an angular velocity of 11.0 rad/s.

Answer:

385 rad

Explanation:

The expression for the angular acceleration of a disk that is in contact with a spinning wheel can be given as:

$\alpha = \frac{\delta \omega}{\delta t}$

where $\delta\omega$ = $\omega_f - \omega_i$

$\alpha = \frac{ \omega_f-\omega_i}{\delta t}$

$\alpha = \frac{ 4.0 rad/s-0 rad/s}{35}$

$\alpha =0.14 rad/s^2$

Angular displacement of a disk can be calculated by using the formula:

$\theta = \omega t$

substituting 11.0 rad/s for $\omega$ and t = 35 s ; we have:

$\theta = 11.0 rad/s * 35 s$

$\theta = 385 rad$

4 0

3 years ago

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A 1-meter-long wire consists of an inner copper core with a radius of 1.0 mm and an outer aluminum sheathe, which is 1.0 mm thic

Mazyrski [523]

Answer:

The total resistance of the wire is = $1.917\times10^{-3}$

Explanation:

Since the wires will both be in contact with the voltage source at the same time and the current flows along in their length-wise direction, the two wires will be considered to be in parallel.

Hence, for resistances in parallel, the total resistance, $R_{Total}$

$\frac{1}{R_{Total}} =\frac{1}{R_{cu} }+\frac{1}{R_{al}}$

Parameters given:

Length of wire = 1 m

Cross sectional area of copper $A_{cu}= \pi r^{2}= \pi \times (1\times 10^{-3} )^{2} =3.142\times10^{-6} m^{2}$

Cross sectional area of aluminium wire

$A_{al}= \pi( R^{2}-r^{2})\\\\ = \pi \times [ (2\times 10^{-3} )^{2}-(1\times 10^{-3} )^{2}] =9.42\times10^{-6} m^{2}\\$

Resistivity of copper $\rho _{cu}=1.7\times 10^{-8} \Omega .m$

Resistivity of Aluminium $\rho _{al}=2.8\times 10^{-8} \Omega .m$

Resistance of copper $R_{cu}= \frac{\rho_{cu} \times l}{A_{cu} } =\frac{1.7\times 10^{-8} \times 1}{3.142\times10^{-6} } =5.41\times 10^{-3}\Omega$

Resistance of aluminium $R_{al}= \frac{\rho_{al} \times l}{A_{al} } =\frac{2.8\times 10^{-8} \times 1}{9.42\times10^{-6} } =2.97\times 10^{-3}\Omega$

The total resistance of the wire can be obtained as follows;

$\frac{1}{R_{Total}} =\frac{1}{5.41\times10^{-3} }+\frac{1}{2.97\times10^{-3}}=521.52\frac{1}{\Omega}$

$R_{Total}= 1.917\times 10^{-3}\Omega$

∴ The total resistance of the wire = $1.917\times 10^{-3}\Omega$

4 0

3 years ago

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What is the mass of a plastic spoon

m_a_m_a [10]

The mass of a plastic spoon

3 0

3 years ago

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How are magnetic poles and electrical charges similar? A. Their same forces repel each other, and their opposite forces attract

melomori [17]

Answer:

A. Their same forces repel each other, and their opposite forces attract each other.

Explanation:

As we know by the property of electric charge that similar charges repel each other and opposite charges attracts each other

So here we have similar situation in magnets also

Two similar poles or like poles of magnet i.e. North - North poles or South - South poles always repel each other

And also we can see that opposite poles of magnet each other i.e. North pole and south pole always attract each other.

SO here correct answer is

A. Their same forces repel each other, and their opposite forces attract each other.

8 0

3 years ago

Pressure that increases with depth in a swimming pool is called ______________ pressure.

lukranit [14]

Hydrostatic pressure.

4 0

4 years ago