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AlekseyPX
3 years ago
14

Which of these discoveries is generally attributed to kepler?

Physics
1 answer:
Ne4ueva [31]3 years ago
4 0
The question is incomplete and no options are given but the answer is;
The orbit of Mars is an ellipse,<span> is generally attributed to Kepler.
Kepler's first law states that; "</span><span>The orbits of the planets are ellipses, with the Sun at one focus of the ellipse.
</span>The planet at that point takes after and follow the ellipse in its orbit, which implies that the distance between Earth and Sun remove is continually changing as the planet goes around its orbit.
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In the periodic table, the most reactive metals are found a. in Group 1, the first column on the left. b. in Period 1, the first
Gnesinka [82]

Answer:

Bottom left corner for whatever group that is

Lithium, sodium, and potassium all react with water

3 0
3 years ago
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Which statement best describes beaches? Beaches change over time as waves move further up the shore. Beaches are always changing
NemiM [27]

Answer:

I think the 1st statement is right.

Explanation:

Wind patterns doesn't stay the same.

Waves don't follow the same patterns.

Waves move further up the shore.

I didn't hear about "waves adding" before..so i guess 1st statement is right.

3 0
3 years ago
What are the steps in order to get the correct outcome?
In-s [12.5K]

1 Purpose/Question. Ask a question. 2 Research. Conduct background research. ... 3 Hypothesis. Propose a hypothesis. ... 4 Experiment. Design and perform an experiment to test your hypothesis. ... 5 Data/Analysis. Record observations and analyze the meaning of the data. ... 6 Conclusion. Conclude whether to accept or reject your hypothesis.

Sure hope this helps you. Pls mark me as brainiest

7 0
2 years ago
Two observers are 300 ft apart on opposite sides of a flagpole. The angles of elevation from the observers to the top of the pol
valentina_108 [34]

Answer:

h = 48.077 ft

Explanation:

given,

distance between two observer = 300 ft

angle of elevation to top pole = 16° and 20°

height of the flagpole = ?

now,

Let h be the height of the flagpole

Let x be the distance of the pole

tan 16^0 = \dfrac{h}{x}

x =\dfrac{h}{tan 16^0}

now,

again applying

tan 20^0 = \dfrac{h}{300-x}

300-x=\dfrac{h}{tan 20^0}

300-3.49 h=2.75 h

6.24h = 300

h = 48.077 ft

3 0
3 years ago
There are only two charged particles in a particular region. Particle 1 carries a charge of 3q and is located on the negative x-
andrezito [222]

Answer:

The net field will be the sum of the fields created by each charge.

where the charge Q in a position r' is given by:

E(r) = k*Q/(r - r')^2

Where k is a constant, and r is the point where we are calculating the electric field.

Then for the charge 3q, in the position r₁ = (-d, 0, 0) the electric field will be:

E₁(r) = k*3q/(r - r₁)^2

While for the other charge of -2q in the position r₂ = (d, 0, 0)

The electric field is:

E₂(r) = -k*2*q/(r - r₂)^2

Then the net field at the point r is:

E(r) = E₁(r) + E₂(r) = k*3q/(r - r₁)^2 + -k*2*q/(r - r₂)^2

E(r) = k*q*( 3/(r - r₁)^2 - 2/(r - r₂)^2)

Then if the we want to find the points r = (x, y, z) such that:

E(r) = 0 = k*q*( 3/(r - r₁)^2 - -k*2*q/(r - r₂)^2)

Then we must have:

0 = ( 3/(r - r₁)^2 - 2/(r - r₂)^2)

Also remember that the distance between two points:

(x, y, z) and (x', y', z') is given by:

D = √( (x - x')^2 + (y - y)^2 + (z -z')^2)

Then we can rewrite:

r - r₁ = √( (x - (-d))^2 + (y - 0 )^2 + (z -0)^2)

       = √( (x + d))^2 + y^2 + z^2)

and

r - r₂ =  √( (x - d)^2 + (y - 0 )^2 + (z -0)^2)

       = √( (x - d))^2 + y^2 + z^2)

Replacing that in our equation we get:

0 = ( 3/(√( (x + d))^2 + y^2 + z^2))^2 - -k*2*q/(√( (x - d))^2 + y^2 + z^2))^2)

0 = (3/((x + d))^2 + y^2 + z^2) - 2/ (x - d))^2 + y^2 + z^2)

We want to find the values of x, y, z such that the above equation is true.

2/ (x - d))^2 + y^2 + z^2) = (3/((x + d))^2 + y^2 + z^2)

2*[((x + d))^2 + y^2 + z^2] = 3*[(x - d))^2 + y^2 + z^2]

2*(x + d)^2  + 2*y^2 + 2*z^2 = 3*(x - d)^2 + 3*y^2 + 3*z^2

2*(x + d)^2 - 3*(x - d)^2 =  3*y^2 + 3*z^2 -  2*y^2 - 2*z^2

2*(x + d)^2 - 3*(x - d)^2  = y^2 + z^2

2*x^2 + 2*2*x*d + 2*d^2 -  3*x^2 + 3*2*x*d - 3*d^2 = y^2 + z^2

-x^2 + 10*x*d - d^2 = y^2 + z^2

we can rewrite this as:

- ( x^2 - 10*x*d + d^2) =  y^2 + z^2

now we can add and subtract 24*d^2 inside the parenthesis to get

- ( x^2 - 10*x*d + d^2 + 24*d^2 - 24*d^2) =  y^2 + z^2

-( x^2 - 2*x*(5d) + 25d^2 - 24d^2) = y^2 + z^2

-(x^2 - 2*x*(5d) + (5*d)^2) + 24d^2 = y^2 + z^2

The thing inside the parenthesis is a perfect square:

-(x - 5d)^2 + 24d^2 = y^2 + z^2

we can rewrite this as:

24d^2 = y^2 + z^2 + (x - 5d)^2

This equation gives us the points (x, y, z) such that the electric field is zero.

Where we need to replace two of these values to find the other, for example, if y = z = 0

24d^2 = (x - 5d)^2

√(24d^2)  = x - 5d

√24*d = x - 5d

√24*d + 5d = x

so in the point (√24*d + 5d, 0, 0) the net field is zero.

7 0
3 years ago
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