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3 years ago

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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What happens to a liquid when it loses energy

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kiruha [24]

Answer:

The speed of Susan is 2.37 m/s

Explanation:

To visualize better this problem, we need to draw a free body diagram.

the work is defined as:

$W=F*d*cos(\theta)$

here we have the work done by Paul and the friction force, so:

$W_p=F_p*d*cos(0)\\F_p=30N*cos(30^o)=26N\\W_p=26*3*(1)=78J$

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Now the change of energy is:

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3 years ago

Read 2 more answers

Martha was leaning out of the window on the second floor of her house and speaking to Steve. Suddenly, her glasses slipped from

DiKsa [7]

Answer:

s = 23.72 m

v = 21.56 m/s²

Explanation:

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time to reach the ground (t) = 2.2 second

we know that

a) s = u t + 0.5 g t²

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g = 9.8 m/s²

s = 0 + 0.5 × 9.8 × 2.2²

s = 23.72 m

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v = √(2× 9.8 × 23.72)

v = √464.912

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3 years ago

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

Margaret [11]

Hi there!

$\boxed{\omega = 0.38 rad/sec}$

We can use the conservation of angular momentum to solve.

$\large\boxed{L_i = L_f}$

Recall the equation for angular momentum:

$L = I\omega$

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

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

$I_m$ = moment of inertia of the merry-go-round (kgm²)

$\omega_m$ = 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:

$I = MR^2$

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

$\omega = \frac{v}{r}$

$L_b = MR^2(\frac{v}{R}) = MRv$

Plug in the given values:

$L_b = (20)(3)(5) = 300 kgm^2/s$

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

$I = MR^2\\I = 20(3^2) = 180 kgm^2$

Use the above equation for conservation of momentum:

$600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}$

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garik1379 [7]

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