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3 years ago

What keeps us from flying into the sky? Why?

Mathematics

2 answers:

mash [69]3 years ago

5 0

Answer:

Gravity.

Step-by-step explanation:

Gravity is what holds the planets in orbit around the sun and what keeps the moon in orbit around Earth. The gravitational pull of the moon pulls the seas towards it, causing the ocean tides. Gravity creates stars and planets by pulling together the material from which they are made.

ElenaW [278]3 years ago

4 0

This is actually a physics problem, but the force of Gravity keeps us from flying into the sky. Gravity is the force by which a planet or other body draws objects toward its center. This force gives us a tendency to remain on the ground and not to float off into the atmosphere.

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Please help me I just don't understand!!!!!!!!!!!!!

Marina86 [1]

Answer:

240pi

Step-by-step explanation:

4 0

3 years ago

Find the velocity and position vectors of a particle that has the given acceleration and the given initial velocity and position

stira [4]

Answer:

$v(t) = (2t+1) \mathbb{i} +3t^2 \mathbb{j} +4t^3 \mathbb{k}$

$r(t) = (t^2+t) \mathbb{i} +(t^3+2) \mathbb{j} +(t^4 - 3) \mathbb{k}$

Step-by-step explanation:

The velocity vector is the integral of the acceleration vector i.e.

$v(t) = \int a(t) dt$

$v(t) = \int (2 \mathbb{i}+6t \mathbb{j} 12t^2 \mathbb{k}) dt$

$v(t) = 2t \mathbb{i}+3t^2 \mathbb{j} 4t^3 \mathbb{k} + C_1$

When $t=0$ , $v(0) = \mathbb{i}$ . Inserting these values in $v(t)$ ,

$C_1= \mathbb{i}$

$v(t) = 2t \mathbb{i}+3t^2 \mathbb{j} 4t^3 \mathbb{k} + \mathbb{i}$

$v(t) = (2t+1) \mathbb{i}+3t^2 \mathbb{j} 4t^3 \mathbb{k}$

The position vector is the integral of the velocity vector i.e.

$r(t) = \int v(t) dt$

$r(t) = \int ((2t+1) \mathbb{i}+3t^2 \mathbb{j} 4t^3 \mathbb{k}) dt$

$r(t) = (t^2+t) \mathbb{i}+t^3 \mathbb{j} t^4 \mathbb{k} + C_2$

When $t=0$ , $r(t) =2\mathbb{j}-3\mathbb{k}$ . Inserting these values in $r(t)$ ,

$C_2=2\mathbb{j}-3\mathbb{k}$

$r(t)= (t^2+t) \mathbb{i}+t^3 \mathbb{j}+ t^4 \mathbb{k} + 2\mathbb{j}-3\mathbb{k}$

$r(t) = (t^2+t) \mathbb{i}+(t^3+2) \mathbb{j}+ (t^4-3) \mathbb{k}$

5 0

4 years ago

Jasmine used the number line to find the distance between 0 and 5. What was Jasmine’s error? Jasmine should have counted from 5

Jobisdone [24]

Answer:

The answer is "Option C".

Step-by-step explanation:

In this question, the distance value is -5 that's why option c is correct.

5 0

3 years ago

The radius of the Earth is approximately 3,960 miles. Find the approximate surface-area-to-volume ratio of the Earth.

Jobisdone [24]

To put it another way, that ratio exists, no matter what distance units you use to express lengths, such as the radius of the Earth, but using different units will result in a different numerical part.
You can't, for instance, say the the radius of the Earth is 3960, and leave it at that. 
If someone comes along who's measuring everything in km, he'll tell you that it's 6373. 
Or if he's using meters, he'll say it's 6,373,000. 
Or in yards, 6,969,600. 

So r = 3960 mi 
And as others have said, the area, A, and volume, V, of a sphere, in terms of its radius, are 
A = 4πr² 
V = 4πr³/3 
so that the area to volume ratio is 
A/V = 3/r 

So the answer is 
3/(3960 mi) = (1/1320) /mi.

5 0

3 years ago

A car is moving at a fixed speed. The distance D that the car covers is then directly proportional to the time T it moves.

worty [1.4K]

We were told that the distance $D$ is directly proportional to the time $T$ .