Which of the following explains why the Earth orbits the sun rather than the sun orbiting the Earth?

AlexFokin [52]

Answer:

Cools ; size

Explanation:

The rate at which magma cools determines the size of the crystals in the new rock. Igneous rocks are formed from the cooling and solidification of molten magma which finds its way to the surface or depth of very low pressure beneath the surface. This place or depth of cooling of magma affects the cooling rate and hence the size of the crystals formed. Igneous rocks formed at depths below the surface have more time to cool and allows more time for Crystal growth and hence produce coarse grained crystal grains called Intrusive igneous rocks which have significantly larger crystals than those formed on the surface which cools rapidly and allowing very little time for crystal growth giving rise to the formation of fine grained crystals and are called extrusive igneous rocks.

6 0

3 years ago

what equastion do you use to solve Riders in a carnival ride stand with their backs against the wall of a circular room of diame

Hitman42 [59]

Answer:

μsmín = 0.1

Explanation:

There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
This friction force has a maximum value, that can be written as follows:

$F_{frmax} = \mu_{s} *F_{n} (1)$

where μs is the coefficient of static friction, and Fn is the normal force,

perpendicular to the wall and aiming to the center of rotation.

This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
This force has the following general expression:

$F_{c} = m* \omega^{2} * r (2)$

where ω is the angular velocity of the riders, and r the distance to the

center of rotation (the radius of the circle), and m the mass of the

riders.

Since Fc is actually Fn, we can replace the right side of (2) in (1), as

follows:

$F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)$

When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

$m* g = m* \mu_{smin} * \omega^{2} * r (4)$

(The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)

Cancelling the masses on both sides of (4), we get:

$g = \mu_{smin} * \omega^{2} * r (5)$

Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

$60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)$

Replacing by the givens in (5), we can solve for μsmín, as follows:

$\mu_{smin} = \frac{g}{\omega^{2} *r} = \frac{9.8m/s2}{(6.28rad/sec)^{2} *2.5 m} =0.1 (7)$

5 0

3 years ago

How long a star lives and what it becomes at the end of its life depends primarly on what

liraira [26]

I believe the answer would be mass. Low mass stars and medium mass stars often become white dwarfs when they die while high mass stars explode in violent explosions called supernovas and usually leave behind a black hole or a neutron star.

4 0

3 years ago

A 0.400-kg object is swung in a circular path and in a vertical plane on a 0.500-m-length string. If the angular speed at the bo

Talja [164]

Answer:

T = 16.72 N

Explanation:

When the object is swung in a circular path, and in a vertical plane, there are two forces external to the object acting on it at any time: the gravity (which is always downward) and the tension in the string (which always points towards the center of the circle).

At the bottom of the circle, the tension is directly upward, so these two forces, are opposite each other, and the difference between them is the centripetal force , which at this point, keeps the object swinging in a circle.

This is the point of the trajectory where T is maximum.

We can apply Newton's 2nd Law, choosing an axis vertical (y-axis) being the upward direction the positive one, as follows:

T- m*g = m*a

The acceleration, at the bottom of the circle, is only normal (as there are no forces in the horizontal direction) , and is equal to the centripetal acceleration, as follows:

ac = v² / r = ω²*r⇒ T- m*g = m*ω²*r

Replacing by the givens, we can solve for T as follows:

T = m* (ω²*r+g) = 0.4 kg*((8.00)² rad/sec²*0.5m)+9.8 m/s²) = 16.72 N

5 0

3 years ago

A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15Â° below the

Kobotan [32]

Answer:

Acceleration of the crate is 0.362 m/s^2.

Explanation:

Given:

Mass of the box, m = 40 kg