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Over [174]
3 years ago
13

The type of brightness in which all

Physics
1 answer:
Gnoma [55]3 years ago
3 0
The answer is C . obvious brightness
You might be interested in
CUAL DE LAS SIGUIENTES LEYES NO PERTENECEN A LOS 3 PRINCIPIOS DE NEWTON? 1.LEY DE LA ATRACCION 2.LEY FUNDAMENTAL DE LA DINAMICA
vodka [1.7K]

Answer:

La Ley de atracción no pertenece a los tres principios de Newton.

Explanation:

Las leyes de Newton son tres principios que sirven para describir el movimiento de los cuerpos, basados en un sistema de referencias inerciales (es aquél que está en reposo o se mueve con velocidad constante, es decir, no tiene aceleración).

Las 3 leyes de Newton son:

  • <em> Primera Ley o Ley de Inercia :</em> Esta ley enuncia que todo cuerpo permanece en su estado de reposo o de movimiento rectilíneo uniforme a menos que otros cuerpos actúen sobre él. Explica que para modificar el estado de movimiento de un cuerpo es necesario actuar sobre él.
  • <em>Segunda ley o Principio Fundamental de la Dinámica </em>: Esta ley dice que la fuerza que actúa sobre un cuerpo es directamente proporcional a su aceleración.  Es decir, si sobre un cuerpo actúa una fuerza, dicho cuerpo modificará su velocidad, es decir, adquirirá una aceleración. Dicha fuerza y su aceleración generada son proporcionales y están relacionadas mediante:  F=m*a donde F es la fuerza aplicada, m es la masa del cuerpo sobre el que se aplica la fuerza y a es la aceleración adquirida por dicho cuerpo.
  • <em>Tercera ley o Principio de acción-reacción </em>: Esta ley expresa que cuando un cuerpo ejerce una fuerza sobre otro, éste ejerce sobre el primero una fuerza igual y de sentido opuesto.

Entonces, <u><em>la Ley de atracción no pertenece a los tres principios de Newton.</em></u>

5 0
3 years ago
An object having a mass of 11.0 g and a charge of 8.00 ✕ 10-5 C is placed in an electric field E with Ex = 5.70 ✕ 103 N/C, Ey =
dezoksy [38]

Answer : F_{x} = 45.6\times10^{-2}\ N, F_{y} = 3040\times10^{-5} N and F_{z} = 0\ N

Explanation :

Given that,

Charge of the object q = 8.00\times 10^{-5}\ C

Electric field in x-direction E_{x} = 5.70\times10^{3}\ N/C

Electric field in y- direction E_{y} = 380\ N/C

Electric field in z - direction E_{z} = 0

Now, using formula

F = q E

Now, the force on the object in x- direction

F_{x} = 8.00\times10^{-5} C \times5.70\times10^{3} N/C

F_{x} = 45.6\times10^{-2}\ N

The force on the object in y- direction

F_{y} = 8.00\times10^{-5}\ C\times 380\ N/C

F_{y} = 3040\times10^{-5} N

The force on the object in z- direction

F_{z} = 8.00\times10^{-5}\ C\times 0

F_{z} = 0\ N

Hence, this is the required solution.




8 0
4 years ago
Explain what the ionosphere is and how it interacts with some radio waves.
kumpel [21]

Answer:

Explained

Explanation:

  • Ionosphere is the outermost layer of the Earth's atmosphere,extending from 60 km to 1,000 km altitude.
  • It mainly consists of ionized gases by the solar radiation.
  • It is very important layer to Earth because it influences the radio propagation to distant places on Earth.
  • Being ionized it reflects high frequency(shortwave) radio waves to the sky as well as to the Earth by a technique called "skip" or "skywave" propagation for the communication over long distances.
7 0
3 years ago
Purse-seine fishing is known to be the safest fishing method for protecting dolphins.
Citrus2011 [14]
<span>Purse-seine fishing is known to be the safest fishing method for protecting dolphins.

This is false. This type of fishing </span><span>method has been most blamed for the unintentional death of dolphins.</span>
4 0
4 years ago
Read 2 more answers
Two metal plates 15mm apart have a potential difference of 750v between them. The force on a small charged sphere placed between
Romashka [77]

Answer:

50,000 V/m

Explanation:

The electric field between two charged metal plates is uniform.

The relationship between potential difference and electric field strength for a uniform field is given by the equation

\Delta V=Ed

where

\Delta V is the potential difference

E is the magnitude of the electric field

d is the  distance between the plates

In this problem, we have:

\Delta V=750 V is the potential difference between the plates

d = 15 mm = 0.015 m is the distance between the plates

Therefore, rearranging the equation we find the strength of the electric field:

E=\frac{\Delta V}{d}=\frac{750}{0.015}=50,000 V/m

6 0
4 years ago
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