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ryzh [129]
3 years ago
15

Explain why an object’s weight is dependent upon where in the universe it is located.

Physics
1 answer:
strojnjashka [21]3 years ago
6 0

Answer:

The weight of any object will depend on its location in the universe, commanded by the law of gravitacion universal de Newton

Explanation:

Everything obeys the law of universal gravity proposed by Newton. Where the attraction force with which one body attracts another depends on the mass of the body that attracts the body with less mass, the radius of the body with the greatest mass and the universal gravitational constant.

For a better understanding of this concept let us use examples with numeric values. In the first example we will determine with what force planet Earth attracts a person of 70 [kg].

And in the second example we will perform the same exercise but on a planet like Jupiter

Example 1:

A person with a mass of 70 [kg] is located on planet Earth which has a mass of 5.97x10^24 [kg] and a terrestrial radius of 6371 [km]. Find the force exerted by the planet upon the person.

We have the  law of universal gravity

F=G*\frac{m_{1} *m_{2} }{r^{2} } \\where\\G = 6.67*10^-11[\frac{N*m^{2} }{kg^{2}} ]\\m_{1}=70[kg]\\m_{2}=5.97*10^{24} [kg]\\r= 6371[km]

Now replacing the values in the equation we have:

F=6.67*10^{-11} *\frac{70*5.97*10^{24} }{(6371*10^3)^{2} } \\F=687[N]

We can appreciate that if we only use the terms of mass of the earth, the gravitational constant and the radius of the earth, we will have the value of the gravity accelaration of the earth. Let's check

g_{earth} =6.67*10^{-11} *\frac{5.97*10^{24} }{(6371*10^3)^{2} } \\g_{earth} = 9.81 [m/s^{2} ]

Example 2:

A person with a mass of 70 [kg] is located on planet Jupiter which has a mass of 1.899x10^27 [kg] and a  radius of 71492 [km]. Find the force exerted by the planet upon the person.

F=6.67*10^{-11}*\frac{70*1.899*10^{27} }{(71492*10^3)^{2} }  \\F= 1734.73[N]

We will calculate the value of the gravity accelaration of Jupiter. Let's check

g_{jupiter} = 6.67*10^{-11} *\frac{1.899*10^{27} }{(71492*10^3)^{2} } \\g_{jupiter}= 24.8 [m/s^2]\\

Therefore an object in jupiter will weigh more than 2.5 times its weight than on planet Earth.

We found that the force of attraction or the weight of any object will depend on its location in the universe, commanded by the law of gravitacion universal de Newton

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Gennadij [26K]

density tomes volume

620x75 ... 3100x15

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The ph of an acidic substance ia between what values on the ph scale
aalyn [17]

0 - 6 is acidic

7 is neutral

8 - 14 is basic

7 0
3 years ago
A train pulls the boxcars. What are the vertical forces present? (1 point)
Greeley [361]

The vertical forces present in a boxcars being pulled by a train are weight and pull of air on the boxcars.

The forces exerted on an object can be resolved into <em>horizontal</em> and <em>vertical components.</em>

<em />

The horizontal components of the forces on an object being pulled include the following;

  • The frictional force on the object; which tends to reduce the motion,
  • force dragging the object forward
  • if the object is inclined to an angle, the horizontal component of the weight

This horizontal force is written as;

\Sigma F_x = 0\\\\F_x - F_k = ma

The vertical component of the force on the object include the following;

  • the weight of the object acting downwards
  • the pull of air on the object acting upwards

The vertical force on the object is written as;

\Sigma F_n = 0\\\\F_n - W = 0

Thus, the vertical forces present in a boxcars being pulled by a train are weight and pull of air on the boxcars.

Learn more here: brainly.com/question/2000189

5 0
3 years ago
Una carga positiva de 4 x 10-5 C, se encuentra a 0.05 m de otra carga positiva de 2 x 10-5 C. Calcular la fuerza que se ejerce e
KATRIN_1 [288]

Answer:

La fuerza que se ejerce entre las dos cargas es 2880 N.

Explanation:

La ley de Coulomb indica que los cuerpos cargados sufren una fuerza atractiva o repulsiva al acercarse. La fuerza es atractiva si las cargas son del signo opuesto y repulsión si son del mismo signo. El valor de la fuerza es proporcional al producto del valor de sus cargas e inversamente proporcional al cuadrado de la distancia que los separa. Esto se expresa matemáticamente como:

F=k*\frac{Q*q}{r^{2} }

donde:

  • F es la fuerza eléctrica de atracción o repulsión. Se mide en Newtons (N).
  • Q y q son los valores de las dos cargas puntuales. Se miden en culombios (C).
  • r es el valor de la distancia que los separa. Se mide en metros (m).
  • k es una constante de proporcionalidad llamada constante de la ley de Coulomb.

En este caso:

  • F= ?
  • Q= 4*10⁻⁵ C
  • q= 2*10⁻⁵ C
  • r= 0.05 m
  • k= 9*10⁹ \frac{N*m^{2} }{C^{2} }

Reemplazando:

F=9*10^{9} \frac{N*m^{2} }{C^{2} }*\frac{4*10^{-5} C*2*10^{-5}C }{(0.05 m)^{2} }

F= 2880 N

<u><em>La fuerza que se ejerce entre las dos cargas es 2880 N.</em></u>

7 0
3 years ago
A moving electron passes near the nucleus of a gold atom, which contains 79 protons and 118 neutrons. At a particular moment the
AURORKA [14]

Answer:

a) 22.471 x 10^-11 N

b) -22.471 x 10^-11 N

Explanation:

given the charge on an electron = 1.6 x 10^-19 C

nucleus contain 79 protons

hence, charge on nucleus =79 x 1.6 x 10^-19 C

distance r = 9 x 10^-9 m

a) therefore, force exerted by the gold nucleus on the electron = Kqe/r^2

= 9 x 10^9 x 79 x 1.6 x 10^-19 x 1.6 x 10^-19 / ( 9 x 10^-9 m)^2

= 22.471 x 10^-11 N

b) force exerted by the electron on the gold nucleus ; = -Force exerted by gold on electron ( From newton third law of motion)

= - 22.471 x 10^-11 N

4 0
3 years ago
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