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Finger [1]
2 years ago
9

A uniform cylindrical rod is 0.63m long and it has cross sectional area of 0.1m. calculate the depth of immersion

Physics
1 answer:
Stels [109]2 years ago
8 0

The depth of immersion of the uniform cylindrical rod is 0.63m.

<h3>What is Depth of immersion?</h3>

This is defined as the distance in which a body reaches when it is submerged.

Depth of immersion = Volume/Cross-sectional area

Volume = Cross-sectional area × length

               = 0.1m² × 0.63m = 0.063m³

Depth of immersion = 0.063m³ / 0.1m²

                                  = 0.63m

Read more about Depth of immersion here brainly.com/question/13554650

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Moment of inertia: A dumbbell-shaped object is composed by two equal masses, m, connected by a rod of negligible mass and length
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We are given that

Mass of each object=m

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I_1=Moment of inertia of the dumbbell shaped object with respect to an axis passing through the center of the rod and perpendicular to it.

I_2=Moment of inertia with respect to an axis passing through one of the masses.

I_1=\frac{mr^2}{4}+\frac{mr^2}{4}

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In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu
Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass 1.67 \times 10^(-27)kg, charge +e = +1.60 \times 10^(-19) C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50 \times 10^6 m/s. The proton comes momentarily to rest at a distance 5.31 \times 10^(-13) m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 5.31 \times 10^{-13} m apart?

Explanation:

The given data is as follows.

Mass of proton = 1.67 \times 10^{-27} kg

Charge of proton = 1.6 \times 10^{-19} C

Speed of proton = 2.50 \times 10^{6} m/s

Distance traveled = 5.31 \times 10^{-13} m

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

  (K.E + P.E)_{initial} = (K.E + P.E)_{final}

 (\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)

where,    \frac{kq_{p}q_{t}}{r} = U = Electric potential energy

     U = (\frac{1}{2}m_{p}v^{2}_{p})

Putting the given values into the above formula as follows.

        U = (\frac{1}{2}m_{p}v^{2}_{p})

            = (\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})

            = 5.218 \times 10^{-15} J

Therefore, we can conclude that the electric potential energy of the proton and nucleus is 5.218 \times 10^{-15} J.

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