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viktelen [127]
3 years ago
10

A thin rod (length = 2.97 m) is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge.

The mass of the rod may be ignored, compared to the mass of the object fixed to the top of the rod. The rod, starting from rest, tips over and rotates downward. (a) What is the angular speed of the rod just before it strikes the floor? (Hint: Consider using the principle of conservation of mechanical energy.)(b) What is the magnitude of the angular acceleration of the rod just before it strikes the floor?
Physics
1 answer:
nlexa [21]3 years ago
3 0

Answer:

a)  w = 2.57 rad / s , b)   α = 3.3 rad / s²

Explanation:

a) Let's use the conservation of mechanical energy, we will write it in two points the highest and when touching the ground

Initial. Higher

       Em₀ = U = m g h

Final. Touching the ground

       Em_{f} = K = ½ I w²

How energy is conserved

       Em₀ = Em_{f}

       mg h = ½ I w2

The moment of specific object inertia

        I = m L²

We replace

       m g h = ½ (mL²) w²

       w² = 2g h / L²

The height of the object is the length of the bar

        h = L

        w = √ 2g / L

       w = √ (2 9.8 / 2.97)

       w = 2.57 rad / s

b) the angular acceleration can be found from Newton's second rotational law

       τ = I α

       W L = I α

       mg L = (m L²) α

       α = g / L

       α = 9.8 / 2.97

       α = 3.3 rad / s²

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Determine the binding energy per nucleon of an mg-24 nucleus. the mg-24 nucleus has a mass of 24.30506. a proton has a mass of 1
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The mass of Mg-24 is 24.30506 amu, it contains 12 protons and 12 neutrons.

Theoretical mass of Mg-24:

The theoretical mass of Mg-24 is:

Hydrogen atom mass = 12 × 1.00728 amu = 12.0874 amu

Neutron mass = 12 x 1.008665 amu = 12.104 amu

Theoretical mass = Hydrogen atom mass + Neutron mass = 24.1913 amu

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Mass defect = Actual mass - Theoretical mass : 24.30506 amu- 24.1913 amu= 0.11376 amu

Calculating the binding energy per nucleon:

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Does crystallization break down rocks at earths surface to form sediment
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A 40W lamp wastes 34 J of energy every second by heating its surroundings.
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Answer:

15\%.

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The efficiency of a machine is the percentage of energy input that was turned into useful energy.

The power rating of this lamp is 40\; \rm W (same as 40\; \rm J \cdot s^{-1},) meaning that 40\; \rm J of energy is supplied to this lamp every second.

The question states that 34\; \rm J out of that 40\; \rm J of energy input would be turned into heat, which is not useful energy output in this scenario. Assuming that all other forms of energy loss is negligible. The rest of the (40\; \rm J - 34\; \rm J) = 6\; \rm J of energy supplied to this lamp would be turned into useful energy output.

Thus, every second, this lamp would receive 40\; \rm J of energy input and would outputs 6\; \rm J of useful work. The efficiency of this lamp would be:

\begin{aligned}& \text{Efficiency} \\ =\; & \frac{\text{Useful energy out}}{\text{Total energy in}} \times 100\% \\ =\; & \frac{6\; \rm J}{40\; \rm J} \times 100\%\\ =\; &15\% \end{aligned}.

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