Based on the law of conversation of energy how can we reasonably improve a machines ability to do work?

Amy throws a softball through the air. What are the different forces acting on the ball while it’s in the air?

aniked [119]

An applied force is a force that is applied to an object by a person or another object.
An attractive force is a force of an attraction (where object are attracted by each other). Gravitation is an example of attractive force.
Normal force is the component, perpendicular to the surface (surface being a plane) of contact.
The softball experiences an applied force as a result of Amy’s throw. As the ball moves, it experiences attractive force from the air it passes through. It also experiences a downward pull because of the normal force.
Solution A.

4 0

3 years ago

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Why are black holes considered the "destroyers and creators" of life? A. Black holes play a role in the formation of galaxies th

bezimeni [28]

Answer:

A

Explanation:

A is just the most logical. Sagittarius doesn't mean anything here.

Light does not originate anywhere. DNA is not formed in a black hole.

6 0

3 years ago

16. Compared to your weight and mass on Earth, if you were on the moon: *

Natalija [7]

Answer:

The answer is D.

Explanation:

There is no gravity in Space so that means that it will decrease your weight but not your mass.

<h2>Please give Brainiest</h2>

5 0

4 years ago

How does the diameter of the disk of milky way galaxy compare to its thickness?

weqwewe [10]

The diameter is about 100 times as great as the thickness.

7 0

4 years ago

At what distance from Earth does the force of gravity exerted by Earth on the coasting spacecraft cancel the force of gravity ex

telo118 [61]

Answer:

346 * 10⁶ m

Explanation:

The force of gravity of the earth that will cancel the the force of gravity exerted by the moon will be equal to each other

Let $F_{e}$ be the force of gravity exerted by the earth

and let $F_{m}$ be the force of gravity exerted by the moon

According to Newton's law of universal gravitation, the force of attraction between two different masses, m₁ and m₂ separated by a distance, d, is given by:

$F = \frac{Gm_{1} m_{2} }{d^{2} }$

Mass of the earth, $m_{e} = 5.97 * 10^{24} kg$

Mass of the moon, $m_{m} = 7.348 * 10^{22} kg$

Mass of the satellite, $m_{s} = ?$

$F_{e} = \frac{G*5.97 * 10^{24} M }{d^{2} }$ ...............................(1)

The earth and the moon are separated by a distance, 3.844 * 10⁸ m

$F_{m} = \frac{G*7.348 * 10^{22} M }{(3.844 * 10^{8} - d) ^{2} }$ ............................(2)

Equating equations (1) and (2)

$\frac{5.97 * 10^{24} }{d^{2} } = \frac{7.348 * 10^{24} }{(3.844* 10^{8} -d)^{2} }$

$(5.97 * 10^{24})(14.78 * 10^{16} -7.688*10^{8}d + d^{2}) = 7.348 * 10^{24} d^{2} \\88.24*10^{40} - 45.9 * 10^{32}d + 5.97 * 10^{24}d^{2} = 7.348 * 10^{24} d^{2}\\ 1.378 * 10^{24}d^{2} + 45.9 * 10^{32}d + 88.24*10^{40} = 0\\$

Factorising out $10^{24}$

$1.378d^{2} + 45.9 * 10^{8}d + 88.24*10^{16} = 0$

Solving for d in the quadratic equation above:

d = 346 * 10⁶ m

4 0

3 years ago