You walk to the corner store at 2 m/s for a total time of 76 seconds. What is the distance to the store?

How did Dalton think atoms formed a new substance?

vivado [14]

Dalton's atomic<span> theory proposed that all matter was composed of </span>atoms<span>, indivisible and indestructible building blocks. While all </span>atoms<span> of an element were identical, different elements had </span>atoms<span> of differing size and mass</span>

7 0

3 years ago

Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars

zhuklara [117]

Kepler's third law is used to determine the relationship between the orbital period of a planet and the radius of the planet.

The distance of the earth from the sun is $1.50 \times 10^{11}\;\rm m$ .

<h3>What is Kepler's third law?</h3>

Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the radius of their orbits. It means that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit.

$T^2 \propto R^3$

Given that Mars’s orbital period T is 687 days, and Mars’s distance from the Sun R is 2.279 × 10^11 m.

By using Kepler's third law, this can be written as,

$T^2 \propto R^3$

$T^2 = kR^3$

Substituting the values, we get the value of constant k for mars.

$687^2 = k\times (2.279 \times 10^{11})^3$

$k = 3.92 \times 10^{-29}$

The value of constant k is the same for Earth as well, also we know that the orbital period for Earth is 365 days. So the R is calculated as given below.

$365^3 = 3.92\times 10^{-29} R^3$

$R^3 = 3.39 \times 10^{33}$

$R= 1.50 \times 10^{11}\;\rm m$

Hence we can conclude that the distance of the earth from the sun is $1.50 \times 10^{11}\;\rm m$ .

To know more about Kepler's third law, follow the link given below.

brainly.com/question/7783290.

6 0

3 years ago

A baseball diamond is a square (don’t be fooled that it is usually shown rotated by 45°). Each side of the square is 90 feet lon

gayaneshka [121]

A line from second to home creates a right triangle with 45 degree angles.

This makes the diagonal line the side length of the square multiplied by the sqrt(2)

Second to home = 90sqrt(2) = 127.279 feet ( round answer as needed)

8 0

3 years ago

An object 1 of mass m1 is separated by some distance d from an object 2 of mass 2m1 . An object 3 of mass m3 is to be placed bet

Harrizon [31]

If the potential energy of the three-object system is to be a maximum (closest to zero), should object 3 be placed closer to object 1, closer to object 2, or halfway between them?

Object 3 should be placed closer to object 1.

Object 3 should be placed on a halfway between object 2 and object 1.

Object 3 should be placed closer to object 2.

Solution

I think that Object 3 should be placed closer to object 2.

6 0

3 years ago

Calculate the magnitude of the effort in the given

Otrada [13]

Moment = Force x Distance from pívot
500x0.5=2.5 x effort
250= 2.5 x effort
effort = 250/2.5= 100N

8 0

3 years ago