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hodyreva [135]
2 years ago
14

How do unbalanced forces acting on an object affect its motion when the object is at rest? What if it is moving?

Physics
2 answers:
inn [45]2 years ago
8 0

Answer:

It pushes it because an unbalanced force is pushing more newtons than something that isn't even moving. Even if it is moving, it depends which side is pushing/pulling the most force.

Montano1993 [528]2 years ago
4 0

Answer:

It pushes it because an unbalanced force is pushing more newtons than something that isn't even moving. Even if it is moving, it depends which side is pushing/pulling the most force.

Explanation:

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Can someone pls help ASAP:(
diamong [38]
I’m not sure but I think it’s
△ m=5 and △= -3 and so

Answer: 5/△-3 m/s

So sorry if it’s wrong
6 0
3 years ago
Planet 1 orbits Star 1 and Planet 2 orbits Star 2 in circular orbits of the same radius. However, the orbital period of Planet 1
hichkok12 [17]

Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

\sum F=ma

We know that the force between two bodies due to gravity is given by the following equation:

F_{g} = G\frac{m_{1}m_{2}}{r^{2}}

in this case we will call:

M= mass of the star

m= mass of the planet

r = distance between the star and the planet

G= constant of gravitation.

so:

F_{g} =G\frac{Mm}{r^{2}}

Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

F_{c}=ma_{c}

where the centripetal acceleration is given by:

a_{c}=\omega ^{2}r

where

\omega = \frac{2\pi}{T}

Where T is the period, and \omega is the angular speed of the planet, so:

a_{c} = ( \frac{2\pi}{T})^{2}r

or:

a_{c}=\frac{4\pi^{2}r}{T^{2}}

so:

F_{c}=m(\frac{4\pi^{2}r}{T^{2}})

so now we can do the sum of forces:

\sum F=ma

F_{g}=ma_{c}

G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})

in this case we can get rid of the mass of the planet, so we get:

G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})

we can now solve this for T^{2} so we get:

T^{2} = \frac{4\pi ^{2}r^{3}}{GM}

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

T_{1}^{2}>T_{2}^{2}

So let's see what's going on there, we'll call:

M_{1}= mass of Star 1

M_{2}= mass of Star 2

So:

\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}

we can get rid of all the constants so we end up with:

\frac{1}{M_{1}}>\frac{1}{M_{2}}

and let's flip the inequality, so we get:

M_{2}>M_{1}

This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.

6 0
2 years ago
Prentice Hall. All
iVinArrow [24]

Answer:

Decreases/Reduces

Explanation:

Fill in the blank:

Consider the equation Work = Force X Distance.

<em>If a machine  increases the distance over which a force is exerted, the force </em>

<em>required to do a given amount of work</em> .........

If the work is a constant value, then by isolating force from the equation, we get:

Force = Work / Distance

By increasing the value of the Distance, then the quotient Work. Distance diminishes, and therefore the required force decreases (diminishes, reduces)

Answer: Decreases/Reduces

7 0
3 years ago
What are dimensionless quantities??​
bearhunter [10]

Answer:

Characteristic numbers are dimensionless numbers used in fluid dynamics to describe a character of the flow. To compare a real situation with a small-scale model it is necessary to keep the important characteristic numbers the same. Names of these numbers were standardized in ISO 31, part 12.

Explanation:

7 0
2 years ago
Charina says that when waves interact with an object, they will interfere with the object, and when waves interact with other wa
Fofino [41]
I would not agree with her since reflection and refraction happens only when waves hit an object. When, waves meet it is either it experiences constructive or destructive interference. Hope this answers the question. Have a nice day.
3 0
2 years ago
Read 2 more answers
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