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4 years ago

12

a meter stick is hung from a string tied at the 25cm mark. A 0.50-kg object is hung From the zero end of the meter stick, and th

e meterstick is balanced horizontally. What is the mass of the meterstick? (a) 0.25kg, (b) 0.5kg (c) 0.75kg (d) 1.0 kg (e) 2.0kg (f) impossible to determine

2 answers:

Vesna [10]4 years ago

6 0

Impossible to determine

pychu [463]4 years ago

3 0

Let m₁ = 0.25m and m₂ = 0.75m, where m is the mass of the meterstick:

The sum of all torques must be zero:
τ = 12.5 * m₁ + 25 * 0.5 - 37.5 * m₂ = 0

Replace m₁ and m₂:
12.5 * 0.25m + 25 * 0.5 = 37.5 * 0,75m

Solve for m:
3.125m + 12.5 = 28.125
25m = 12.5
m = 0.5

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Answer:

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So the radius of its orbit is

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For the ion of oxygen-18, we have:

$m_B = \frac{18}{16}m_A = 2.99\cdot 10^{-26}kg$

$q_B = 1.6\cdot 10^{-19}C$ (it is singly charged)

$v_B=2.90\cdot 10^6 m/s$

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$r_B=\frac{m_B v_B}{q_B B_B}=\frac{(2.99\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.417 m$

After each ion has travelled a semicircle, the separation between the two ions will be twice the difference in their radius, so:

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3 years ago

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saveliy_v [14]

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3 years ago

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velikii [3]

Answer: A

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3 years ago

The wave function of a particle is exp(i(kx-omegat)), where x is distance, t is time, and k and co are positive real numbers. Th

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Answer:

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4 years ago