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Angelina_Jolie [31]
3 years ago
8

Two metallic rods A and B of different materials have same length. The linear expansivity of A is 12×10–6 oC–1and cubical expans

ivity of B is 24×10–6 oC–1. If both rods are heated from 20oC to 100oC, the length of rod A will be....................... The length of rod B.
Physics
1 answer:
uysha [10]3 years ago
6 0

Answer:

The length of rod A will be <u>greater than </u>the length of rod B

Explanation:

We, know that the formula for final length in linear thermal expansion of a rod is:

L' = L(1 + ∝ΔT)

where,

L' = Final Length

L = Initial Length

∝ = Co-efficient of linear expansion

ΔT = Change in temperature

Since, the rods here have same original length and the temperature difference is same as well. Therefore, the final length will only depend upon the coefficient of linear expansion.

For Rod A:

∝₁ = 12 x 10⁻⁶ °C⁻¹

For Rod B:

∝₂ = β₂/3

where,

β₂ = Coefficient of volumetric expansion for rod B = 24 x 10⁻⁶ °C⁻¹

Therefore,

∝₂ = 24 x 10⁻⁶ °C⁻¹/3

∝₂ = 8 x 10⁻⁶ °C⁻¹

Since,

∝₁ > ∝₂

Therefore,

L₁ > L₂

So, the length of rod A will be <u>greater than </u>the length of rod B

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This question is incomplete, the complete question is;

The electric force due to a uniform external electric field causes a torque of magnitude 20.0 × 10⁻⁹ N⋅m on an electric dipole oriented at 30° from the direction of the external field. The dipole moment of the dipole is 7.5 × 10⁻¹² C⋅m.

What is the magnitude of the external electric field?

If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle?

Answer:

- the magnitude of the external electric field is 5333.3 N/C

- the magnitude of the charge on each particle is 3.0 × 10⁻¹² C  ≈ 3 nC

Explanation:

Given that;

Torque = 20.0 × 10⁻⁹ N⋅m

dipole moment = 7.5 × 10⁻¹²

∅ = 30°

The moment T of restoring couple is;

T = PEsin∅

E = T/Psin∅

we substitute

E = 20.0 × 10⁻⁹ N⋅m / (7.5 × 10⁻¹²) sin(30°)

E = 20.0 × 10⁻⁹ / 3.75 × 10⁻¹²

E =  5333.3 N/C

Therefore, the magnitude of the external electric field is 5333.3 N/C

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p = ql

q = p / l

given that l = 2.5 mm = 0.0025 m

we substitute

q = 7.5 × 10⁻¹² / 0.0025

q = 3.0 × 10⁻¹² C ≈ 3 nC

Therefore, the magnitude of the charge on each particle is 3.0 × 10⁻¹² C ≈ 3 nC

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3 years ago
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Answer:

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Apply Newton's second law:

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On a safety test course, a 1000 kg car heading north at 5 m/s collides head-on with an 800 kg car heading south at 4 m/s. At the
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Answer:

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Explanation:

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p_f=p_i

The total momentum is the sum of the momentum of each of the bodies we're dealing, in our case the moment of each car, then:

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It's important to note that when we talk about momentum and velocity direction matters, so we're are going to choose a system of reference where quantities pointing north are positive and pointing south are negative. So, the initial velocity of 1000 kg car is vni=5 m/s, initial velocity of 800 kg car is vsi=-4 m/s and the final velocity of 1000 kg car is vnf=-1 m/s. Now we can solve (1) for vsf and use the values we already have:

v_{sf}=\frac{m_nv_{ni}+m_sv_{si}-m_nv_{nf}}{m_s}=\frac{(1000)(5)+(800)(-4)-(1000)(-1)}{800}

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Because the sign is positive the direction is to heading north.

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