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2 years ago

In Astronomy List the systems of which the Earth is part of, from smallest to largest (most inclusive)?

Physics

1 answer:

djverab [1.8K]2 years ago

3 0

Answer:

In order from largest to smallest these terms are:

In order from largest to smallest these terms are:Universe
In order from largest to smallest these terms are:UniverseGalaxy
In order from largest to smallest these terms are:UniverseGalaxySolar system
In order from largest to smallest these terms are:UniverseGalaxySolar systemStar
In order from largest to smallest these terms are:UniverseGalaxySolar systemStarPlanet
In order from largest to smallest these terms are:UniverseGalaxySolar systemStarPlanetMoon
In order from largest to smallest these terms are:UniverseGalaxySolar systemStarPlanetMoonAsteroid

In order from largest to smallest these terms are:UniverseGalaxySolar systemStarPlanetMoonAsteroidWe can generally say that a Moon is larger than an asteroid however some moons are quite large while others are smaller than many known asteroids. Asteroids are small rocky bodies that orbit the Sun, while moons are objects that orbit planets. Jupiter and Saturn have attracted many objects out of the asteroid belt and gathered them as moons.

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A constant net force is applied to an object. Suddenly, the object's mass doubles, but the net force remains the same. For this

____ [38]

D: Half

We know that,

F = m.a, where F is the force, m is the object's mass and a is the acceleration.

In the first case, we observed that a1 = F/m.

In the second case, we observed that the mass has been doubled, so a2 = F/2m .

By the ratio of the two cases, we get

a1/a2 = F/m / F/2m

or, a1/a2 = 2

or, a1 = 2.a2

or, a1/2 = a2

Therefore, the acceleration gets half of it's original measurement.

8 0

3 years ago

The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .

ASHA 777 [7]

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is $W=1.20 \cdot 10^{-3}J$

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: $W_e=q\Delta V$

where

q is the charge

$\Delta V$ is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

$K_f - K_i = W + W_e = W+q\Delta V$

and since the charge starts from rest, $K_i = 0$ , so the formula becomes

$K_f = W+q\Delta V$

In this problem, we have

$W=1.20 \cdot 10^{-3}J$ is the work done by the external force

$q=-6.70 \mu C=-6.7\cdot 10^{-6}C$ is the charge

$K_f = 4.72\cdot 10^{-4}J$ is the final kinetic energy

Solving the formula for $\Delta V$ , we find

$\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V$

4 0

3 years ago

Read 2 more answers

A horizontal pipe narrows from a radius of 0.250 m to 0.1000 m. If the speed of the water in the pipe is 1.00 m/s in the larger-

umka21 [38]

Answer:

6.25 m/s

196 kg/s

Explanation:

The areas of the pipe from big to small:

$A = \pi R^2 = \pi*0.25^2 = 0.196 m^2$

$a = \pi r^2 = \pi*0.1^2=0.0314m^2$

As the product of speed and cross-section area is constant, the speed in the smaller pipe would be

$AV = av$

$v = \frac{AV}{a} = \frac{0.196 * 1}{0.0314} = 6.25 m/s$

The mass flow rate would be:

$\dot{m} = \ro AV = 1000 * 0.196 * 1 = 196 kg/s$

4 0

3 years ago

Every chemical element goes through natural exponential decay, which means that over time its atoms fall apart. The speed of eac

Naddik [55]

Answer:

t = (ti)ln(Ai/At)/ln(2)

t = 14ln(16)/ln(2)

Solving for t

t = 14×4 = 56 seconds

Explanation:

Let Ai represent the initial amount and At represent the final amount of beryllium-11 remaining after time t

At = Ai/2^n ..... 1

Where n is the number of half-life that have passed.

n = t/half-life

Half life = 14

n = t/14

At = Ai/2^(t/14)

From equation 1.

2^n = Ai/At

Taking the natural logarithm of both sides;

nln(2) = ln(Ai/At)

n = ln(Ai/At)/ln(2)

Since n = t/14

t/14 = ln(Ai/At)/ln(2)

t = 14ln(Ai/At)/ln(2)

Ai = 800

At = 50

t = 14ln(800/50)/ln(2)

t = 14ln(16)/ln(2)

Solving for t

t = 14×4 = 56 seconds

Let half life = ti

t = (ti)ln(Ai/At)/ln(2)

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4 years ago

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Calculate the force between two objects that have masses of 75. Kilograms and 2,100. Kilograms. Their centers of gravity are sep

scoundrel [369]

Answer:

use the formula above it's easy

3 0

2 years ago

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In Astronomy List the systems of which the Earth is part of, from smallest to largest (most inclusive)?​

In Astronomy List the systems of which the Earth is part of, from smallest to largest (most inclusive)?