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MrRissso [65]
3 years ago
11

Why are objects that fall near Earth's surface rarely in free fall?

Physics
1 answer:
sveta [45]3 years ago
6 0

Answer: Air exerts forces on falling objects near Earth's surface.

Explanation: Objects falling near the earth surface are rarely in free fall due to the force exerted by air on falling object near the earth( air resistance).

The acceleration of free fall tend to pull the body towards the earth surface while air resistance (drag) tends to act in the opposite direction.

Given that the weight of the body is always  constant.

The drag acts in the upward direction, thereby negating the downward weight of the object.

From Newton's Second law:

Force = mass × acceleration due to gravity

Acceleration = force / mass

The net force acting on the object becomes :

(Downward weight - upward drag)

This hampers the free fall of the object due to gravity.

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3 years ago
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How do I solve this arithmetic sequence?
Serjik [45]

Answer:

The 16ᵗʰ term of this sequence is 82

Step-by-step explanation:

Here,

First Term = a₁ = 9

Common Difference = (d) = 2

Now, For 16ᵗʰ term, n = 16

<em>aₙ = a + (n - 1)d</em>

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4 0
2 years ago
A small object with momentum 7.0 kg∙m/s approaches head-on a large object at rest. The small object bounces straight back with a
EastWind [94]

Answer:

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

Explanation:

Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:

p_{S,1}+p_{B,1} = p_{S,2}+p_{B,2} (1)

Where:

p_{S,1}, p_{S,2} - Initial and final momemtums of the small object, measured in kilogram-meters per second.

p_{B,1}, p_{B,2} - Initial and final momentums of the big object, measured in kilogram-meters per second.

If we know that p_{S,1} = 7\,\frac{kg\cdot m}{s}, p_{B,1} = 0\,\frac{kg\cdot m}{s} and p_{S, 2} = 4\,\frac{kg\cdot m}{s}, then the final momentum of the big object is:

7\,\frac{kg\cdot m}{s} + 0\,\frac{kg\cdot m}{s} = 4\,\frac{kg\cdot m}{s}+p_{B,2}

p_{B,2} = 3\,\frac{kg\cdot m}{s}

The magnitude of the large object's momentum change is:

p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}-0\,\frac{kg\cdot m}{s}

p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

4 0
2 years ago
The angular separation of two stars is 0.1 arcseconds and you photograph them with a telescope that has an angular resolution of
gizmo_the_mogwai [7]

Answer:

Will see them as only one star

Explanation:

Solution:

- The angular resolution of a telescope means the minimum quantity that can be visualized. Since their angular separation ( 0.1 arcseconds ) is smaller than the telescope's angular resolution (1 arcseconds ), your photograph will seem to show only one star rather than two.

7 0
2 years ago
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