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devlian [24]
3 years ago
6

Which elements have similar behavior? barium silicon aluminum strontium osmium beryllium

Physics
2 answers:
tresset_1 [31]3 years ago
8 0
Which elements have similar behavior? 
Barium, strontium, beryllium. 

Ilya [14]3 years ago
3 0

<u>Answer:</u> The elements which have similar behavior are Barium, strontium and beryllium.

<u>Explanation:</u>

Periodic table is divided into 7 periods and 18 groups.

The elements belonging to the same group will show chemical properties. Groups are the vertical columns in the periodic table.

From the given options:

Barium, strontium and beryllium, all belong to the group 2.

Silicon belong to group 14.

Aluminium belong to group 13 and Osmium belong to group 8.

Hence, the elements which have similar behavior are Barium, strontium and beryllium.

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An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o
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Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

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In case of simple harmonic motion, the time period of oscillation is given by :

T=2\pi\sqrt{\dfrac{m}{k}}

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m=\dfrac{2.45}{9.8}

m = 0.25 kg

k=\dfrac{4\pi^2m}{T^2}

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A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the
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\boxed{\omega = 0.38 rad/sec}

We can use the conservation of angular momentum to solve.

\large\boxed{L_i = L_f}

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We can begin by writing out the scenario as a conservation of angular momentum:

I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)

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\omega_m = angular velocity of merry go round (rad/sec)

\omega_f = final angular velocity of COMBINED objects (rad/sec)

I_b = moment of inertia of boy (kgm²)

\omega_b= angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

I = MR^2

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Plug in the given values:

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Now, we must solve for the boy's moment of inertia:

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