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vampirchik [111]

2 years ago

8

A neutron star is as dense as Group of answer choices water the nucleus of an atom the center of the Earth our astronomy textboo

k a white dwarf star

1 answer:

AVprozaik [17]2 years ago

4 0

A neutron star is considered as dense as the nucleus of an atom, and it represents a special type of star.

<h3>What is a neutron star?</h3>

A neutron star is a special kind of star generated after the death of a giant star that produces a supernova.

These stars (neutron stars) have a diameter really small, e.g., an area of ten kilometers in diameter.

In conclusion, a neutron star is as dense as the nucleus of an atom, and it represents a special type of star.

Learn more about neutron stars here:

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Two shuffleboard disks of equal mass, one orange and the other yellow, are involved in an elastic, glancing collision. The yello

denis-greek [22]

Answer:

The final speeds of the yellow and orange shuffleboard disks are 3.015 meters per second and 3.989 meters per second, respectively.

Explanation:

In this case, we see a two-dimension system formed by two shuffleboard disks, which do not experiment any effect from external forces, so that the Principle of Linear Momentum can be applied. The equations of equilibrium are respectively:

$m_{Y}\cdot \vec v_{Y,o}+m_{O}\cdot \vec v_{O,o} = m_{Y}\cdot \vec v_{Y}+m_{O}\cdot \vec v_{O}$ (Eq. 1)

Where:

$m_{Y}$ , $m_{O}$ - Masses of the yellow and orange shuffleboard disks, measured in kilograms.

$\vec v_{Y,o}$ , $\vec v_{O,o}$ - Initial vectors velocity of the yellow and orange shuffleboard disks, measured in meters per second.

$\vec v_{Y}$ , $\vec v_{O}$ - Final vectors velocity of the yellow and orange shuffleboard disks, measured in meters per second.

If we know that $m_{Y} = m_{O}$ , then the system is now reduced into this form:

$\vec v_{Y,o}+\vec v_{O,o} = \vec v_{Y}+\vec v_{O}$ (Eq. 2)

Given that $\vec v_{Y,o} = \left(0\,\frac{m}{s}, 0\,\frac{m}{s}\right)$ , $\vec v_{O,o} = \left(5\,\frac{m}{s}, 0\,\frac{m}{s} \right)$ , $\vec v_{Y} = v_{Y}\cdot \left(\cos 52.92^{\circ},-\sin 52.92^{\circ}\right)$ and $\vec v_{O} = v_{O}\cdot (\cos 37.08^{\circ},\sin 37.08^{\circ})$ , then we find these two equations of equilibrium:

x-Direction:

$v_{Y}\cdot \cos 52.92^{\circ}+v_{O}\cdot \cos 37.08^{\circ} = 5\,\frac{m}{s}$ (Eq. 3)

y-Direction:

$-v_{Y}\cdot \sin 52.92^{\circ}+v_{O}\cdot \sin 37.08^{\circ} = 0\,\frac{m}{s}$ (Eq. 4)

The solution of this system of linear equations is:

$v_{Y} = 3.015\,\frac{m}{s}$ and $v_{O} = 3.989\,\frac{m}{s}$

The final speeds of the yellow and orange shuffleboard disks are 3.015 meters per second and 3.989 meters per second, respectively.

3 0

3 years ago

An explosive projectile is launched straight upward to a maximum height h. At its peak, it explodes, scattering particles in all

Varvara68 [4.7K]

Answer:

θ = tan⁻¹ ( $\frac{19.6 \ h}{v}$ )

Explanation:

This problem must be solved using projectile launch ratios. Let's analyze the situation, the projectile explodes at the highest point, therefore we fear the height (i = h), the speed at this point is the same, but the direction changes, we are asked to find the smallest angle of the speed in the point of arrival with respect to the x-axis.

The speed at the arrival point (y = 0)

v² = vₓ² + v_y²

Let's see how this angle changes, for two extreme values:

* The particle that falls from the point of explosion, in this case the speed is vertical

v = v_y

the angle with the horizontal is 90º

* The particle leaves horizontally from the point of the explosion, the initial velocity is horizontal

vₓ = v

the final velocity for y = 0

v_f = vₓ² + v_y²

therefore the angle has a value greater than zero and less than 90º

As they ask for the smallest angle, we can see that we must solve the last case

the output velocity is horizontal vₓ = v

Let's find the velocity when it hits the ground y = 0, with y₀ = h

$v_{y}^2$ = $v_{oy}^2$ - 2 g (y-y₀)

v_{y}^2 = - 2g (0- y₀)

let's calculate

v_{y}^2 = 2 9.8 h

we use trigonometry to find the angle

tan θ = $\frac{v_y}{v_x}$

θ = tan⁻¹ ( $\frac{v_y}{v_x}$ )

let's calculate

θ = tan⁻¹ ( $\frac{19.6 \ h}{v}$ )

3 0

3 years ago

(DUE IN FIVE MINUTES, QUICK)

goldfiish [28.3K]

Answer:

Well you would be flouting out in space so it will change your weight a lot because you can't stop moving.

8 0

3 years ago

Read 2 more answers

What is air pressure?

Dmitry_Shevchenko [17]

Air Pressure is a force that is exerted onto your body by air molecules.

3 0

3 years ago

. An object has a position given by ~r(t) = [3.0 m − (4.00 m/s)t]ˆı + [6.0 m − (8.00 m/s2 )t 2 ]ˆ , where all quantities are in

kupik [55]

Answer:

(c) 16 m/s²

Explanation:

The position is $r(t) = [3.0 \text{ m} - (4.00 \text{ m/s})t]\hat{i} + [6.0 \text{m} - (8.00 \text{ m/s}^2 )t^2 ]\hat{j}$ .

The velocity is the first time-derivative of <em>r(t).</em>

<em /> $v(t) = \dfrac{d}{dt}r(t) = -4.00\,\hat{i} -16t\,\hat{j}$ <em />

The acceleration is the first time-derivative of the velocity.

$a(t) = \dfrac{d}{dt} v(t) = -16\hat{j}$

Since <em>a(t)</em> does not have the variable <em>t</em>, it is constant. Hence, at any time,

$a = -16\hat{j}$

Its magnitude is 16 m/s².

4 0

3 years ago