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2 years ago

13

The gravitational force between the sun and every object in the solar system helps keep each object in its own unique orbit arou

nd the sun. Lesson 3. 03 Question 2 options: True False.

2 answers:

saw5 [17]2 years ago

8 0

Answer:

True

Explanation:

i did the Quiz

Kazeer [188]2 years ago

3 0

Answer:

A: true

Explanation:

I took the test its right, trust me I got 100%

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Newton first law of motion ?

maw [93]

Answer:

The law of inertia

Explanation:

A body at rest will remain at rest, and a body in motion will remain in motion unless it is acted upon by an external force

8 0

3 years ago

Read 2 more answers

The radius of a planet is 2400 km, and the acceleration due to gravity at its surface is 3.6 m/s2.

kiruha [24]

Answer:

$3.1\cdot10^{23}\:\mathrm{kg}$

Explanation:

We can use Newton's Universal Law of Gravitation to solve this problem:

$g_P=G\frac{m}{r^2}$ ., where $g_P$ is acceleration due to gravity at the planet's surface, $G$ is gravitational constant $6.67\cdot 10^{-11}$ , $m$ is the mass of the planet, and $r$ is the radius of the planet.

Since acceleration due to gravity is given as $m/s^2$ , our radius should be meters. Therefore, convert $2400$ kilometers to meters:

$2400\:\mathrm{km}=2,400,000\:\mathrm{m}$ .

Now plugging in our values, we get:

$3.6=6.67\cdot10^{-11}\frac{m}{(2,400,000)^2}$ ,

Solving for $m$ :

$m=\frac{2,400,000^2\cdot3.6}{6.67\cdot 10^{-11}},\\m=\fbox{$3.1\cdot10^{23}\:\mathrm{kg}$}$ .

6 0

2 years ago

We see a full moon by reflected sunlight. how much earlier did the light that enters our eye leave the sun

galina1969 [7]

We know that the source of light in the universe is the Sun. Hence, the light we see as moonlight travels from the Sun's surface, to the moon, then to Earth. So, before being able to solve this problem, we have to know the distance between the Sun and the moon, and the distance between the moon and Earth. In literature, these values are 3.8×10⁵ km (Sun to moon) and 384,400 km (moon to Earth). Knowing that the speed of light is 300,000 km per second, then the total time would be

Time = distance/speed
Time = (3.8×10⁵ km + 384,400 km)/300,000 km/s
Time = 2.548 seconds

Thus, it only takes 2.548 for the light from the Sun to reach to the Earth as perceived to be what we call moonlight.

7 0

3 years ago

a child riding a bicycle at 15 meters per second accelerates at -3,0 meters per second? for 4.0 seconds. What is the child's spe

Vika [28.1K]

Answer:

<h2> 27m/s</h2>

Explanation:

Given data

initital velocity u=15m/s

deceleration a=3m/s^2

time t= 4 seconds

final velocity v= ?

Applying the expression

v=u+at------1

substituting our data into the expression we have

v=15+3*4

v=15+12

v=27m/s

The velocity after 4 seconds is 27m/s

5 0

2 years ago

The energy levels of a particular quantum object are -11.7 eV, -4.2 eV, and -3.3 eV. If a collection of these objects is bombard

gogolik [260]

To solve this problem it is necessary to apply an energy balance equation in each of the states to assess what their respective relationship is.

By definition the energy balance is simply given by the change between the two states:

$|\Delta E_{ij}| = |E_i-E_j|$

Our states are given by

$E_1 = -11.7eV$

$E_2 = -4.2eV$

$E_3 = -3.3eV$

In this way the energy balance for the states would be given by,

$|\Delta E_{12}| = |E_1-E_2|\\|\Delta E_{12}| = |-11.7-(-4.2)|\\|\Delta E_{12}| = 7.5eV\\$

$|\Delta E_{13}| = |E_1-E_3|\\|\Delta E_{13}| = |-11.7-(-3.3)|\\|\Delta E_{13}| = 8.4eV$

$|\Delta E_{23}| = |E_2-E_3|\\|\Delta E_{23}| = |-4.2-(-3.3)|\\|\Delta E_{23}| = 0.9eV$

Therefore the states of energy would be

Lowest : 0.9eV

Middle :7.5eV

Highest: 8.4eV

8 0

3 years ago