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2 years ago

The gravitational force between two objects increases as

Physics

1 answer:

Misha Larkins [42]2 years ago

8 0

Answer: Their masses increases or the distance between them decreases

Explanation:

According to Newton's law of Gravitation, the gravitational force $F$ exerted between two bodies of masses $m1$ and $m2$ and separated by a distance $r$ is equal to the product of their masses and inversely proportional to the square of the distance:

$F=G\frac{(m1)(m2)}{r^2}$ (1)

Where $G$ is the Gravitational Constant

As we can see, if one body increases its mass or if both bodies increase their mass, the gravitational force will increase as well.

In addition, is the distance between the two bodies is decreased, the gravitational force will increase.

Therefore:

The gravitational force between two objects increases as their mass increases or the distance between them decreases.

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"How did your current and voltage measurements differ between the series and parallel circuits you created

irakobra [83]

Answer:

Series circuit:

The voltage that is measured across the circuit is different.

The current measured in a series circuit remains the same at all points in the circuit.

Parallel circuit:

The current measured across each resistor varies

The voltage measured across a parallel circuit will remain the same

Explanation:

Series and parallel circuits behave differently when it comes to the circulation of current and the interaction with a potential difference.

In a series circuit, the resistances are connected end to end. As a result, the voltage that is measured across the circuit is different once resistance is encountered. However, the current measured in a series circuit remains the same at all points in the circuit.

A parallel circuit behaves in an exactly opposite manner to the series circuit. In a parallel circuit, the resistances are connected side by side. As a result of this, the current measured across each resistor varies as there are circuit branches through which electric current can flow into. On the other hand, the voltage measured across a parallel circuit will remain the same

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2 years ago

A 2 kg object traveling at 5 m s on a frictionless horizontal surface collides head-on with and sticks to a 3 kg object initiall

Svetlanka [38]

Answer: (d)

Explanation:

Given

Mass of object $m=2\ kg$

Speed of object $u=5\ m/s$

Mass of object at rest $M=3\ kg$

Suppose after collision, speed is v

conserving momentum

$\Rightarrow mu+0=(m+M)v\\\\\Rightarrow v=\dfrac{2\times 5}{2+3}\\\\\Rightarrow v=2\ m/s$

Initial kinetic energy

$k_1=\dfrac{1}{2}\times 2\times 5^2\\\\k_1=25\ J$

Final kinetic energy

$k_2=\dfrac{1}{2}\times (2+3)\times 2^2\\\\k_2=10\ J$

So, it is clear there is decrease in kinetic energy . Thus, energy decreases and velocity becomes 2 m/s.

4 0

2 years ago

¿Cuál es el lado positivo sobre el auge de la industrialización

Nesterboy [21]

Answer:

El lado positivo que tuvo el auge de la industrialización es principalmente que: Permitió el desarrollo económico de una gran cantidad de países. Marco un nuevo estilo de vida con mayor índice de globalización y producción. Creó una gran cantidad de fuentes de empleo.

Explanation:

3 0

3 years ago

What is a charge-coupled device (CCD), and how is it used in astronomy?

Sedaia [141]

Answer:

Charge-coupled device (CCD) is a device that receives and transfers an electrical charge to the next region

Explanation:

Charge-coupled device (CCD) is a device that receives and transfers an electrical charge to the next region where it can be modified like changing it to a electronic value.

In astronomy, high-powered telescopes can be used with CCD device image sensor cameras. The imaging system can concentrate for a number of hours on one place in space once the Earth's rotation synchronizes with the telescope.

6 0

3 years ago

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

ch4aika [34]

Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

m = Mass of electron = $9.11\times 10^{-31}\ kg$

B = Magnetic field = 0.044 T

q = Charge of electron = $1.6\times 10^{-19}\ C$

The centripetal force and the magnetic forces are conserved

$m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}$

Velocity of first electron

$v=\frac{Bqr_1}{m}\\\Rightarrow v=\frac{0.044\times 1.6\times 10^{-19}\times 0.01}{9.11\times 10^{-31}}\\\Rightarrow v_1=77277716.79473\ m/s$

Velocity of second electron

$v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s$

Total kinetic energy is given by

$K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J$

Converting to eV

$1\ J=\frac{1}{1.6\times 10^{-19}}\ eV$

$1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV$

The energy of incident electron is 114.92749 keV

5 0

3 years ago

The gravitational force between two objects increases as​

The gravitational force between two objects increases as