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1 year ago

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Two girls,(masses m1 and m2) are on roller skates and stand at rest, close to each other and face-to-face. Girl 1 pushes squarel

y against girl 2 and sends her moving backward. Assuming the girls move freely on their skates, write an expression for the speed with which girl 1 moves. ?

1 answer:

Arturiano [62]1 year ago

5 0

Answer:

$\vec{v}_1 = -\frac{\vec{v}_2m_2}{m_1}$

Explanation:

The center of mass of the system (two girls) is constant, as the velocity of the center of mass of the system is also constant.

$\vec{v}_{cm} = \frac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2}$

<u>The initial velocity of the system is zero, since both girls are at rest.</u> So the velocity of the total system at any point should be zero as well.

$0 = \frac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2}\\\vec{v}_1 = -\frac{\vec{v}_2m_2}{m_1}$

This is true, because there is no friction between the girls and the ground. Otherwise, the velocity of the center of mass wouldn't be constant.

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An engineer uses aluminum to build an airplane rather than composite materials that are lighter and stronger. He does this becau

AleksandrR [38]

Answer:

choosing a material that will show warning before it fails

Explanation:

According to my research on different architectural engineering techniques, I can say that based on the information provided within the question this is an example of choosing a material that will show warning before it fails. By choosing aluminum he can detect certain failures a long time before it actually happens since aluminum shows signs of wear and tear and doesn't just break immediately.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

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2 years ago

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The air in a 6.00 L tank has a pressure of 2.00 atm. What is the final pressure, in atmospheres, when the air is placed in tanks

ser-zykov [4K]

Explanation:

Given that,

Initial volume of tank, V = 6 L

Initial pressure, P = 2 atm

We need to find the final pressure when the air is placed in tanks that have the following volumes if there is no change in temperature and amount of gas:

(a) V' = 1 L

It is a case of Boyle's law. It says that volume is inversely proportional to the pressure at constant temperature. So,

$PV=P'V'\\\\P'=\dfrac{PV}{V'}\\\\P'=\dfrac{6\times 2}{1}\\\\P'=12\ atm$

(b) V' = 2500 mL

New pressure becomes :

$PV=P'V'\\\\P'=\dfrac{PV}{V'}\\\\P'=\dfrac{6\times 2}{2500\times 10^{-3}}\\\\P'=4.8\ atm$

(c) V' = 750 mL

New pressure becomes :

$PV=P'V'\\\\P'=\dfrac{PV}{V'}\\\\P'=\dfrac{6\times 2}{750\times 10^{-3}}\\\\P'=16\ atm$

(d) V' = 8 L

New pressure becomes :

$PV=P'V'\\\\P'=\dfrac{PV}{V'}\\\\P'=\dfrac{6\times 2}{8}\\\\P'=1.5\ atm$

Hence, this is the required solution.

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2 years ago

Scientific knowledge builds upon previous knowledge."

notka56 [123]

Answer:

B

Explanation:

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2 years ago

A turntable that is initially at rest is set in motion with a constant angular acceleration α. What is the magnitude of the angu

bekas [8.4K]

Explanation:

If the turntable starts from rest and is set in motion with a constant angular acceleration α. Let $\omega$ is the angular velocity of the turntable. We know that the rate of change of angular velocity is called the angular acceleration of an object. Its formula is given by :

$\alpha =\dfrac{\omega_f-\omega_i}{t}$

$\alpha =\dfrac{\omega-0}{t}$

$\alpha =\dfrac{\omega}{t}$

$t=\dfrac{\omega}{\alpha }$ ............(1)

Using second equation of kinematics as :

$\theta=\omega_i t+\dfrac{1}{2}\alpha t^2$

$\theta=\dfrac{1}{2}\alpha t^2$

Using equation (1) in above equation

$\theta=\dfrac{1}{2}\times \dfrac{\omega^2}{\alpha }$

In one revolution, $\theta=4\pi$ (in 2 revolutions)

$4\pi =\dfrac{1}{2}\times \dfrac{\omega^2}{\alpha }$

$\omega=\sqrt{8\pi \alpha}$

$\omega=2\sqrt{2\pi \alpha}$

Hence, this is the required solution.

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2 years ago

Read 2 more answers

Suppose that now you want to make a scale model of the solar system using the same ball bearing to represent the sun. How far fr

Dahasolnce [82]

Answer:

d = 0.645 m <em>(assuming a radius of the ball bearing of 3 mm)</em>

Explanation:

<u>The given information is:</u>

<em>The distance from the center of the sun to the center of the earth is 1.496x10¹¹m = $d_{e}$ </em>
<em>The radius of the sun is 6.96x10⁸m = $r_{s}$ </em>

<u>We need to assume a radius for the ball bearing, so suppose that the radius is 3 mm = $r_{b}$ </u>.

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:

$\frac{r_{s}}{r_{b}} = \frac{6.96\cdot 10^{8}m}{3\cdot 10^{-3}m} = 2.32\cdot 10^{11}$

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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1 year ago