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2 years ago

6. A billiard ball traveling at 4.0 m/s has an elastic head-on collision with a billiard ball of equal mass

Physics

1 answer:

Yakvenalex [24]2 years ago

7 0

Answer is C‼️.......................

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Consult Interactive Solution 10.37 to explore a model for solving this problem. A spring is compressed by 0.0647 m and is used t

padilas [110]

Answer:

$\omega=32.14\ rad/s$

Explanation:

Given that,

The compression in the spring, x = 0.0647 m

Speed of the object, v = 2.08 m/s

To find,

Angular frequency of the object.

Solution,

We know that the elation between the amplitude and the angular frequency in SHM is given by :

$v=\omega\times A$

A is the amplitude

In case of spring the compression in the spring is equal to its amplitude

$\omega=\dfrac{v}{A}$

$\omega=\dfrac{2.08\ m/s}{0.0647\ m}$

$\omega=32.14\ rad/s$

So, the angular frequency of the spring is 32.14 rad/s.

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3 years ago

What is the mass, in kg, of a 136 pound gymnast on Earth?

Rzqust [24]

Answer:

61.6886 kg

Explanation:

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2 years ago

What are the primary colors? What are the secondary colors?

Jet001 [13]

Color Basics

Three Primary Colors (Ps): Red, Yellow, Blue.

Three Secondary Colors (S'): Orange, Green, Violet

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2 years ago

Read 2 more answers

such smut. when he asks how this goal can be accomplished he volunteers to preview every film that will be shown in town. If Ada

juin [17]

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2 years ago

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles

belka [17]

Answer:

$F_a=5.67\times 10^{-5}\ N$

$F_b=3.49\times 10^{-5}\ N$

$F_c=9.16\times 10^{-5}\ N$

Explanation:

Given:

mass of particle A, $m_a=363\ kg$

mass of particle B, $m_b=517\ kg$

mass of particle C, $m_c=154\ kg$

All the three particles lie on a straight line.

Distance between particle A and B, $x_{ab}=0.5\ m$

Distance between particle B and C, $x_{bc}=0.25\ m$

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

Force on particle A due to particles B & C:

$F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}$

$F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})$

$F_a=5.67\times 10^{-5}\ N$

Force on particle C due to particles B & A:

$F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}$

$F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )$

$F_c=9.16\times 10^{-5}\ N$

Force on particle B due to particles C & A:

$F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}$

$F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2} )$

$F_b=3.49\times 10^{-5}\ N$

3 0

3 years ago