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3 years ago

When Nina jumps into the air, the Earth pulls her back down through the force of gravity. What else occurs?

Physics

2 answers:

Alex Ar [27]3 years ago

5 0

It is A. hope this helps

Nookie1986 [14]3 years ago

5 0

1) WRONG. Since gravitational force is an attractive force.
2) WRONG. The Earth does accelerate but the acceleration is so small that it's negligible. However it DOES ACCELERATE.
3) CORRECT.
4) WRONG. Mass means gravitational field, so Nina might be small compared to the Earth but it still creates a gravitational field around her. This answer is also wrong because it's breaking Newton's third law of action-reaction. If Earth pulls Nina, then Nina pulls the Earth back with the same force. Yes, with the SAME FORCE, however F=m*a, so the great difference in mass makes Nina's acceleration much bigger than the Earth's one.

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Read 2 more answers

The Cartesian coordinate of a point in the xy plane are (x,y)=(-3.50,-2.50)m. Find the poler coordinate of this point

Masja [62]

Answer:

The polar coordinate of $P(x,y) = (-3.50\,m,-2.50\,m)$ is $P (r,\theta) = (4.301\,m, 215.538^{\circ})$ .

Explanation:

Given a point in rectangular form, that is $P(x,y) = (x,y)$ , its polar form is defined by:

$P(x,y) = (r,\theta)$ (1)

Where:

$r$ - Norm, measured in meters.

$\theta$ - Direction, measured in sexagesimal degrees.

The norm of the point is determined by Pythagorean Theorem:

$r = \sqrt{x^{2}+y^{2}}$ (2)

And direction is calculated by following trigonometric relation:

$\theta = \tan^{-1} \frac{y}{x}$ (3)

If we know that $x = -3.50\,m$ and $y = -2.50\,m$ , then the components of coordinates in polar form is:

$r = \sqrt{(-3.50\,m)^{2}+(-2.50\,m)^{2}}$

$r \approx 4.301\,m$

Since $x < 0\,m$ and $y < 0\,m$ , direction is located at 3rd Quadrant. Given that tangent function has a period of 180º, we find direction by using this formula: