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3 years ago

What two things increase the gravitational pull between two objects?

2 answers:

8 0

Two things that increase the gravitational pull between two objects are decreasing the distance between the objects and increasing the mass of the objects. the force due to gravity is F = GMm/(r^2) where G is the gravitational constant, M and m are the masses of the two objects, and r is the distance between them.

stiv31 [10]3 years ago

5 0

In order to increase the gravitational forces between two objects,
you have exactly two choices:

-- Increase the product of the two masses.

-- Decrease the distance between their centers of mass.

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Mekhanik [1.2K]

Answer:

B because the earth rotational axis tilt away from the sun.

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3 years ago

If the motion of B is uniformly accelerated, at what time will both graphs have

Misha Larkins [42]

Answer:

Im just here for the points man sorry

Explanation:

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3 years ago

At a particular instant, a stationary observer on the ground sees a package falling with speed v1 at an angle to the vertical. t

Lerok [7]

V1 * sin(θ) where θ is the angle v1 makes with the vertical.

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4 years ago

A cylinder of mass mm is free to slide in a vertical tube. The kinetic friction force between the cylinder and the walls of the

Zinaida [17]

Answer:

$y = \frac{-f +/- \sqrt{f^{2} +2kmg}}{k}$

Explanation:

Let y₀ be the initial position of the cylinder when the spring is attached and y its position when it is momentarily at rest.From work-kinetic energy principles, The work done by the spring force + work done by friction + work done by gravity = kinetic energy change of the cylinder

work done by the spring force = ¹/₂k(y₀² - y²)

work done by friction = - f(y - y₀)

work done by gravity = mg(y - y₀)

kinetic energy change of the cylinder = ¹/₂m(v₁² - v₀²)

So ¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = ¹/₂m(v₁² - v₀²)

Since the cylinder starts at rest, v₀ = 0. Also, when it is momentarily at rest, v₁ = 0

¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = ¹/₂m(0² - 0²)

¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = 0

¹/₂ky₀² + fy₀ - mgy₀ -¹/₂ky² - fy + mgy = 0

¹/₂ky₀² + fy₀ - mgy₀ = ¹/₂ky² + fy - mgy

Let y₀ = 0, then the left hand side of the equation equals zero. So,

0 = ¹/₂ky² + fy - mgy

¹/₂ky² + fy - mgy = 0

Using the quadratic formula

$y = \frac{-f +/- \sqrt{f^{2} - 4 X\frac{k}{2} X -mg}}{2 X \frac{k}{2} }\\ y = \frac{-f +/- \sqrt{f^{2} +2kmg}}{k}$

4 0

3 years ago

Astronomers observe two separate solar systems each consisting of a planet orbiting a sun. The two orbits are circular and have

Oxana [17]

Answer:

(D) 3

Explanation:

The angular momentum is given by:

$\vec{L}=\vec{r}\ X \ \vec{p}$

Thus, the magnitude of the angular momenta of both solar systems are given by:

$L_1=Rm_1v_1=Rm_1(\omega R)=R^2m_1(\frac{2\pi}{T_1})=2\pi R^2\frac{m_1}{T_1}\\\\L_2=Rm_2v_2=2\pi R^2\frac{m_2}{T_2}$

where we have taken that both systems has the same radius.

By taking into account that T1=3T2, we have

$L_1=2\pi R^2\frac{m_1}{3T_2}=\frac{1}{3}2\pi R^2\frac{1}{T_2}m_1=\frac{1}{3}\frac{L_2}{m_2}m_1$

but L1=L2=L:

$L=\frac{1}{3}L\frac{m_1}{m_2}\\\\\frac{m_1}{m_2}=3$

Hence, the answer is (D) 3

HOPE THIS HELPS!!

3 0

3 years ago