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2 years ago

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:

Force between object 1 and object 2, F = 18 N

Let the mass of object 1 = m1
Let the mass of object 2 = m2
Let the distance between the two object = r

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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enot [183]

Answer:

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B. $U_1 = \dfrac{\epsilon_0 A V^2}{6d}$

C. $U_2 = \dfrac{K\epsilon_0 A V^2}{2d}$

Explanation:

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$C=\dfrac{\epsilon A}{d}$

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For air or free space, $\epsilon$ is $\epsilon_0$ called the permittivity of free space. In general, $\epsilon=\epsilon_r \epsilon_0$ where $\epsilon_r$ is the relative permittivity or dielectric constant of the material between the plates. It is a factor that determines the strength of the material compared to air. In fact, for air or vacuum, $\epsilon_r=1$ .

The energy stored in a capacitor is the average of the product of its charge and voltage.

$U = \dfrac{QV}{2}$

Its charge, $Q$ , is related to its capacitance by $Q=CV$ (this is the electrical definition of capacitance, a ratio of the charge to its voltage; the previous formula is the geometric definition). Substituting this in the formula for $U$ ,

$U = \dfrac{CV^2}{2}$

A. Substituting for $C$ in $U$ ,

$U_0 = \dfrac{\epsilon_0 A V^2}{2d}$

B. When the distance is $3d$ ,

$U_1 = \dfrac{\epsilon_0 A V^2}{2\times3d}$

$U_1 = \dfrac{\epsilon_0 A V^2}{6d}$

C. When the distance is restored but with a dielectric material of dielectric constant, $K$ , inserted, we have

$U_2 = \dfrac{K\epsilon_0 A V^2}{2d}$

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3 years ago

A cyclist rides 6.2 km east, then 9.28 km in a direction 27.27 degrees west of north, then 7.99 km west. A. What is the magnitud

Pani-rosa [81]

Answer:

Explanation:

given,

cyclist ride 6.2 km east and then 9.28 km in the direction of 27.27° west of north and then 7.99 km west.

vertical component = 9.28 cos∅

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horizontal axis component = 9.28 sin ∅

= 9.28 sin 27.27°

= 4.5 km

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= 7.99 -(6.2-4.5)

= 6.29 km

displacement

$d = \sqrt{6.29^2+8.24^2}$

d = 10.37 km

b) $tan \theta = \dfrac{6.29}{8.24}$

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