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Over [174]
3 years ago
13

"a uniform thin rod of length 0.962 m is hung from a horizontal nail passing through a small hole in the rod located 0.048 m fro

m the rod's end. when the rod is set swinging about the nail at small amplitude, what is the period of oscillation?"
Physics
1 answer:
exis [7]3 years ago
5 0
<span>1.57 seconds. The rod hanging from the nail constructs a physical pendulum. The period of such a pendulum follows the formula T = 2*pi*sqrt(L/g) where T = time L = length of pendulum g = local gravitational acceleration So the problem becomes one of determining L. It's tempting to consider L to be the distance between the center of mass and the pivot, but that isn't the right value. The correct value is the distance between the pivot and the center of percussion. So let's determine what that is. We can treat the uniform thin rod as an uniform beam and for an uniform beam the distance between the center of mass and the center of percussion is expressed as b = L^2/(12A) where b = distance between center of mass and center of percussion L = length of beam A = distance between pivot and center of mass Since the rod is uniform, the CoM will be midway from either end, or 0.962 m / 2 = 0.481 m from the end. The pivot will therefore be 0.481 m - 0.048 m = 0.433 m from the CoM Now let's calculate the distance the CoP will be from the CoM: b = L^2/(12A) b = (0.962 m)^2/(12 * 0.433 m) b = (0.925444 m^2)/(5.196 m) b = 0.178107005 m With the distance between the CoM and CoP known, we can now calculate the effective length of the pendulum. So: 0.433 m + 0.178107005 m = 0.611107005 m And finally, with the effective length known, let's calculate the period. T = 2*pi*sqrt(L/g) T = 2*pi*sqrt((0.611107005 m)/(9.8 m/s^2)) T = 2*pi*sqrt(0.062357858 s^2) T = 2*pi*0.249715554 s T = 1.569009097 s Rounding to 3 significant figures gives 1.57 seconds. Let's check if this result is sane. Looking up "Seconds Pendulum", I get a length of 0.994 meters which is longer than the length of 0.611 meters calculated. But upon looking closer at the "Seconds Pendulum", you'll realize that it's period is actually 2 seconds, or 1 second per swing. So the length of the calculated pendulum is sane.</span>
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Bohr’s atomic model differed from Rutherford's because it explained that electrons exist in specified energy levels surrounding
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Answer:

electrons exist in specified energy levels

Explanation:

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However, Rutherford did not specify anything about the orbits of the electrons. Later, Bohr predicted that the electrons actually orbit the nucleus in specific orbits, each orbit corresponding to a specific energy level. Bohr's model found confirmation in the observation of the emission spectrum lines: when an electron in one of the higher energy level jumps down into an orbit with lower energy, the atom emits a photon which has an energy exactly equal to the difference in energy between the two orbits (and this energy of the photon corresponds to a precise wavelength).

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an apple falling off a tree

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3 years ago
A 0.877 mol sample of N2(g) initially at 298 K and 1.00 atm is held at constant volume while enough heat is applied to raise the
MissTica

Answer : The value of q\text{ and }\Delta U is 286.2 J and 286.2 J respectively.

Explanation : Given,

Moles of sample = 0.877 mol

Change in temperature = 15.7 K

First we have to calculate the heat absorbed by the system.

Formula used :

q=n\times c_v\times \Delta T

where,

q = heat absorbed by the system = ?

n = moles of sample = 0.877 mol

\Delta T = Change in temperature = 15.7 K

c_v = heat capacity at constant volume of N_2 (diatomic molecule) = \frac{5}{2}R

R = gas constant = 8.314 J/mol.K

Now put all the given value in the above formula, we get:

q=0.877mol\times \frac{5}{2}\times 8.314J/mol.K\times 15.7K

q=286.2J

Now we have to calculate the change in internal energy of the system.

\Delta U=q+w

As we know that, work done is zero at constant volume. So,

\Delta U=q=286.2J

Therefore, the value of q\text{ and }\Delta U is 286.2 J and 286.2 J respectively.

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3 years ago
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The net vertical force is 0.0057879 N

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