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3 years ago

What is the basis for rutherford's planetary model?

1 answer:

3 0

The basis for Rutherford's Planetary model, was the results he got from experiments.

He observed that most of the alpha particles he fired at a gold foil, passed through the foil, but only few were deflected back. So he concluded that most of the Atom would be empty space, with a positive entity at the center.

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What part of the gun position equipment supports all of the elevating parts of the gun?

BARSIC [14]

The gunstock it’s also called stock or shoulder stock

5 0

3 years ago

If Judy walks 2 blocks north, 5 blocks west, and then 2 blocks south, what is her final

VladimirAG [237]

Answer: 2 blocks east if i'm not mistaken.

7 0

3 years ago

what is the name of the area around a charged object where the object can exert a force on other charged objects?

shusha [124]

Answer:

An electric field is a region around a charged object where the object's electric force is exerted on other charged objects. Electric fields get weaker the farther away they are from the charge. An electric field is invisible. You can use the field line to represent it.

Explanation:

4 0

2 years ago

Juan's mother drives 12.5 miles southwest to her favorite shopping mall. What is the velocity of her

SVETLANKA909090 [29]

Answer:

Calculate using the formula

Explanation:

velocity= displacement (m)/time(s)

1 mile =1.6km

1km=1000m

7 0

3 years ago

A point charge A of charge +4micro coloumb and another B of -1 micro coloumb are placed at a distance in air 1m apart then the d

andrew11 [14]

Answer:

Explanation:

Given that,

A point charge is placed between two charges

Q1 = 4 μC

Q2 = -1 μC

Distance between the two charges is 1m

We want to find the point when the electric field will be zero.

Electric field can be calculated using

E = kQ/r²

Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.

Then, the magnitude of the electric at point x is zero.

E = kQ1 / r² + kQ2 / r²

0 = kQ1 / x² - kQ2 / (1-x)²

kQ1 / x² = kQ2 / (1-x)²

Divide through by k

Q1 / x² = Q2 / (1-x)²

4μ / x² = 1μ / (1 - x)²

Divide through by μ

4 / x² = 1 / (1-x)²

Cross multiply

4(1-x)² = x²

4(1-2x+x²) = x²

4 - 8x + 4x² = x²

4x² - 8x + 4 - x² = 0

3x² - 8x + 4 = 0

Check attachment for solution of quadratic equation

We found that,

x = 2m or x = ⅔m

So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.

5 0

3 years ago