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Marianna [84]
2 years ago
12

Objects 1 and 2 attract each other with a gravitational force

Physics
1 answer:
Katena32 [7]2 years ago
7 0

The new gravitational force of the two objects will be 432 N.

The given parameters:

  • <em>Force between object 1 and object 2, F = 18 N</em>

  • <em>Let the mass of object 1 = m1</em>
  • <em>Let the mass of object 2 = m2</em>
  • <em>Let the distance between the two object = r</em>

The gravitational force between the two objects is calculated by applying Newton's law of universal gravitation;

F = \frac{Gm_1m_2}{r^2} \\\\G = \frac{Fr^2}{m_1m_2}\\\\G = \frac{Gm_1m_2}{r^2} \\\\G = \frac{18r^2}{m_1m_2}

  • when the mass of the object 1 = 2m1
  • mass of the object 2 = 3m2
  • distance between the two object = ¹/₂r

\frac{18r^2}{m_1 m_2 } = \frac{F_2(\frac{1}{2}r)^2 }{2m_1 \times 3m_2} \\\\\frac{18r^2}{m_1 m_2 } = \frac{F_2 \times r^2}{4 \times 2m_1 \times 3m_2} \\\\\frac{18r^2\times 4 \times 2m_1 \times 3m_2}{m_1 m_2\times r^2 } = F_2\\\\432 \ N = F_2

Thus, the new gravitational force of the two objects will be 432 N.

Learn more about Newton's law of universal gravitation here: brainly.com/question/9373839

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Velocity of the electron at the centre of the ring, v=1.37\times10^7\ \rm m/s

Explanation:

<u>Given:</u>

  • Linear charge density of the ring=0.1\ \rm \mu C/m
  • Radius of the ring R=0.2 m
  • Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C

Potential due the ring at a distance x from the centre of the rings is given by

V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V

Let\Delta U be the change in potential Energy given by

\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s

So the electron will be moving with v=1.37\times10^7\ \rm m/s

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3 years ago
An atom of uranium 238 emits an alpha particle (an atom of He) and recoils with a velocity of 1.895 * 10^ 5 m/sec . With velocit
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<u>Answer:</u> The velocity of released alpha particle is 1.127\times 10^7m/s

<u>Explanation:</u>

According to law of conservation of momentum, momentum can neither be created nor be destroyed until and unless, an external force is applied.

For a system:

m_1v_1=m_2v_2

where,

m_1\text{ and }v_1 = Initial mass and velocity

m_2\text{ and }v_2 = Final mass and velocity

We are given:

m_1=238u\\v_1=1.895\times 10^{5}m/s\\m_2=4u\text{ (Mass of }\alpha \text{ -particle)}\\v_2=?m/s

Putting values in above equation, we get:

238\times 1.895\times 10^5=4\times v_2\\\\v_2=\frac{238\times 1.895\times 10^5}{4}=1.127\times 10^7m/s

Hence, the velocity of released alpha particle is 1.127\times 10^7m/s

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Answer:

A is the answer. Im only 12 and i hope this explanation helps you.

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