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3 years ago

An object has a mass of 10 kilograms and is accelerating at 5 m/s/s, what force pushed the object?

Physics

2 answers:

Karo-lina-s [1.5K]3 years ago

7 0

Answer: 2km/m/s/s

Explanation:

10km divided by 5m/s/s = 2km/m/s/s

Lynna [10]3 years ago

6 0

Answer:

hope helps

Explanation:

In a weightless environment a force of 5 Newtons is applied horizontally to the right on a rock with a mass of 1 kg and to a pebble with a mass of 0.1 kg.

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4. With a diameter that's 11 times larger than Earth's, _______ is the largest planet.

Komok [63]

The answer is B.Jupiter

7 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

3 years ago

Se deja caer una moneda desde cierta altura. Si se desprecian los efectos del aire, ¿cómo varía la fuerza neta sobre la moneda a

forsale [732]

Answer:

Ok, primero pensemos en una situación normal.

La moneda comienza a caer, pero la moneda esta inmersa en una sustancia, el aire. El aire comienza a aplicar una resistencia al movimiento de la moneda, y esta resistencia incremente a medida que la velocidad de la moneda incremente. Llega un punto en el que esta nueva fuerza es igual a la fuerza gravitatoria, y en sentido opuesto, lo que causa que la fuerza neta sea 0, y que la moneda caiga a velocidad constante hasta que esta impacta con el suelo.

Ahora, en este caso tenemos que ignorar los efectos del aire, entonces no hay ninguna fuerza que se oponga a la fuerza gravitatoria, entonces la fuerza neta no cambia a medida que cae (La fuerza neta cambia cuando la moneda impacta el suelo).

También se puede analizar el caso en el que, como la fuerza gravitatoria decrece con el radio al cuadrado, a medida que la moneda cae, la fuerza gravitatoria incrementa. El tema es que en para estas dimensiones, ese cambio en la fuerza gravitacional es imperceptible,

3 0

3 years ago

If you have observed something about the world and want to

ExtremeBDS [4]

Answer: a

Explanation: :)

5 0

2 years ago

Is newton third law accurate for electric forces?

inn [45]

An object with greater charge will exert a greater force on an object than an object with smaller charge would. However, if you consider two charges that exert a force on each other, regardless of the magnitude of charge, both charges will exert an equal force on each other because of Newton's third law.

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2 years ago