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1 year ago

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A 3.0-kg mass and a 5.0-kg mass hang vertically at the opposite ends of a very light rope that goes over an ideal pulley. If the

masses are gently released, what is the resulting acceleration of the masses

1 answer:

AleksAgata [21]1 year ago

5 0

Answer:

acceleration = 2.4525‬ m/s²

Explanation:

Data: Let m1 = 3.0 Kg, m2 = 5.0 Kg, g = 9.81 m/s²

Tension in the rope = T

Sol: m2 > m1

i) for downward motion of m2:

m2 a = m2 g - T

5 a = 5 × 9.81 m/s² - T

⇒ T = 49.05‬ m/s² - 5 a Eqn (a)‬

ii) for upward motion of m1

m a = T - m1 g

3 a = T - 3 × 9.8 m/s²

⇒ T = 3 a + 29.43‬ m/s² Eqn (b)

Equating Eqn (a) and(b)

49.05‬ m/s² - 5 a = T = 3 a + 29.43‬ m/s²

49.05‬ m/s² - 29.43‬ m/s² = 3 a + 5 a

19.62 m/s² = 8 a

⇒ a = 2.4525‬ m/s²

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Explanation:

It is given that,

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Mass of Buffy, $m_2=55\ kg$

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Let $u_1$ is the Buffy's velocity just before the collision. Using the conservation of linear momentum as :

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1 year ago

If you're ever standing on a mountaintop when a dark cloud passes overhead and your hair stands up, get off the mountain fast. H

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Answer:

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Explanation:

While standing on the top of a mountain if a person gets its hairs stand up after a cloud passes over, this might happen due to the static electric charges on the lower surface of the cloud are opposite in nature to that of hairs which the hairs would have acquired by the passing of dry winds which would have resulted in the loss of electron from the hair tip.

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Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consid

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Answer:

a

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b

$z = (1 \ m , 4 \ m )$

Explanation:

From the question we are told that

Their distance apart is $d = 5.00 \ m$

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Let the distance from source A where the construct interference occurred be z

Generally the path difference for constructive interference is

$z - (d-z) = m \lambda$

Now given that we are considering just the straight line (i.e points along the line connecting the two sources ) then the order of the maxima m = 0

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$|z-(d-z)| = (2m + 1)\frac{\lambda}{2}$

=> $|2z - d |= (0 + 1)\frac{\lambda}{2}$

=> $|2z - d| =\frac{\lambda}{2}$

substituting values

$|2z - 5| =\frac{6}{2}$

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So

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and

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Answer:

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E₂ = -σ₂ / 2ε₀

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E_total = (σ₁ + σ₂) / 2ε₀

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The conductive sheet in the middle pate undergoes an induced load that is created by the other two plates, but because the conductive plate the charges are mobile and are replaced.

E_total = (0.51 +0.52) 10⁻⁶ / 2 8.85 10⁻¹²

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Note that the field is independent of the distance between the plates

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1 year ago