PHYSICS HIGH SCHOOL

In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be balanced? The answer is 13.4 In the mobile what is the value for m3 to the nearest hundredth of a kilogram?

Answers

Answer 1

Answer:

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

The first mass is $m_1 = 0.42kg$

The second mass is $m_2 = 0.47kg$

From the question we can see that at equilibrium the moment about the point where the string holding the bar (where $m_1 \ and \ m_2$ are hanged ) is attached is zero

Therefore we can say that

$m_1 * 15cm = m_2 * xcm$

Making x the subject of the formula

$x = (m_1 * 15)/(m_2)$

$= (0.42 * 15)/(0.47)$

$x = 13.4 cm$

Looking at the diagram we can see that the tension T on the string holding the bar where $m_1 \ and \ m_2$ are hanged is as a result of the masses ( $m_1 + m_2$ )

Also at equilibrium the moment about the point where the string holding the bar (where ( $m_1 +m_2$ ) and $m_3$ are hanged ) is attached is zero

So basically

$(m_1 + m_2 ) * 20 = m_3 * 30$

$(0.42 + 0.47) * 20 = 30 * m_3$

Making $m_3$ subject

$m_3 = ((0.42 + 0.47) * 20 )/(30 )$

$m_3 = 0.59 kg$

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Answers

In this exercise we have to use the knowledge in distance, in this way we will find that the proportional distance found is:

$d = 0.645 m$

So from the information given in the text we find that:

The distance from the center of the sun to the center of the earth is $1.496*10^(11) \ m$

The radius of the sun is $6.96*10^(8)m$

We need to assume a radius for the ball bearing, so suppose that the radius is $3 mm$

First, we need to find in what way or manner often the radius of the brightest star exist considerable respect to the range of the ball significance, that exist given apiece following equating:

$(r_a)/(r_b)= \frac{6.96*10^8}{3*10{-3}} =2.32*10^(11)$

Now, we can calculate the distance from the center of the sun to the center of the sphere representing the earth:

$d_(s) = (d_(e))/(r_(s)/r_(b)) = (1.496 \cdot 10^(11) m)/(2.32\cdot 10^(11)) = 0.645 m$

See more about distance at brainly.com/question/989117

Answer:

d = 0.645 m(assuming a radius of the ball bearing of 3 mm)

Explanation:

The given information is:

The distance from the center of the sun to the center of the earth is 1.496x10¹¹m = $d_(e)$
The radius of the sun is 6.96x10⁸m = $r_(s)$

We need to assume a radius for the ball bearing, so suppose that the radius is 3 mm = $r_(b)$ .

First, we need to find how many times the radius of the sun is bigger respect to the radius of the ball bearing, which is given by the following equation:

$(r_(s))/(r_(b)) = (6.96\cdot 10^(8)m)/(3\cdot 10^(-3)m) = 2.32\cdot 10^(11)$

Now, we can calculate the distance from the center of the sun to the center of the sphere representing the earth, $d_(s)$ :

[tex] d_{s} = \frac{d_{e}}{r_{s}/r_{b}} = \frac{1.496 \cdot 10^{11} m}{2.32\cdot 10^{11}} = 0.645 m

I hope it helps you!

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