All categories

3 years ago

How can you make the moon go around in a bigger circle

Physics

2 answers:

zlopas [31]3 years ago

5 0

You would have to give it more mechanical energy.

Like, strap a bunch of powerful rockets to one side of the moon, with all of them pointing in the direction that the moon is already moving in its orbit. Then blast away.

NOTE: There aren't enough rockets or rocket fuel on Earth to make a difference, even if you used ALL of them. The mass of the moon is about

<em>73,476,730,900,000,000,000,000 kilograms</em>

(rounded to the nearest hundred trillion kilograms.)

That's a lot.

Alex_Xolod [135]3 years ago

5 0

I doubt it if you can try

You might be interested in

Select all numbers that are correctly rounded

Andreas93 [3]

A and B i hope it helped

7 0

2 years ago

A person is standing on an elevator initially at rest at the first floor of a high building. The elevator then begins to ascend

GREYUIT [131]

Answer:

The found acceleration in terms of h and t is:

$a=\frac{h}{5(t_1)^2}$

Explanation:

(The complete question is given in the attached picture. We need to find the acceleration in terms of h and t in this question)

We are given 3 stages of movement of elevator. We'll first model them each of the stage one by one to find the height covered in each stage. After that we'll find the total height covered by adding heights covered in each stage, and equate it to Total height h. From that we can find the formula for acceleration.

<h3></h3><h3>Stage 1</h3>

Constant acceleration, starts from rest.

Distance = $y = \frac{1}{2}a(t_1)^2$

Velocity = $v_1=at_1$

<h3>Stage 2</h3>

Constant velocity where

Velocity = $v_o=v_1=at_1$

Distance =

<h3> $y_2=v_2(t_2)\\\text{Where~}t_2=4t_1 ~\text{and}~ v_2=v_1=at_1\\y_2=(at_1)(4t_1)\\y_2=4a(t_1)^2\\$ </h3><h3 /><h3>Stage 3</h3>

Constant deceleration where

Velocity = $v_0=v_1=at_1$

Distance =

$y_3=v_1t_3-\frac{1}{2}a(t_3)^2\\\text{Where}~t_3=t_1\\y_3=v_1t_1-\frac{1}{2}a(t_1)^2\\\text{Where}~ v_1t_1=a(t_1)^2\\y_3=a(t_1)^2-\frac{1}{2}a(t_1)^2\\\text{Subtracting both terms:}\\y_3=\frac{1}{2}a(t_1)^2$

<h3>Total Height</h3>

Total height = y₁ + y₂ + y₃

Total height = $\frac{1}{2}a(t_1)^2+4a(t_1)^2+\frac{1}{2}a(t_1)^2 = 5a(t_1)^2$

<h3 /><h3>Acceleration</h3>

Find acceleration by rearranging the found equation of total height.

Total Height = h

h = 5a(t₁)²

$a=\frac{h}{5(t_1)^2}$

6 0

3 years ago

Select all correct answers....Covalent compounds

yaroslaw [1]

I know for sure that the third one is correct

5 0

2 years ago

Receiver

77julia77 [94]

Answer:

What is the best description for the volume of air volume of air provided in a high quality rescue breath?

Explanation:

A) only Enough air to create a visible rise of the chest.

B) Until you can no longer force air in.

C) plenty of air make sure it is adequate to sustain life

D) clear and obvious rise of the chest, sustained over a few seconds

5 0

2 years ago

What’s an example of contact friction

kotegsom [21]

Answer:

a nightstand on a lamp table

Explanation:

5 0

3 years ago