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ankoles [38]
3 years ago
7

Do electric field lines actually exist?​

Physics
1 answer:
konstantin123 [22]3 years ago
8 0

Answer: No

Explanation:they do not exist electric field lines are only imaginary

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Sound can travel through a medium, but it does not require a medium.<br> a. True.<br> b. False
attashe74 [19]

That's false.  No medium = no sound.
7 0
3 years ago
A brick of mass 5 kg is released from rest at a height of 3 m. How fast is it going when it hits the ground? Acceleration due to
sineoko [7]

Taking into account the definition of kinetic, potencial and mechanical energy, when the brick hits the ground, it has a speed of 7,668 m/s.

<h3>Kinetic energy</h3>

Kinetic energy is a form of energy. It is defined as the energy associated with bodies that are in motion and this energy depends on the mass and speed of the body.

Kinetic energy is defined as the amount of work necessary to accelerate a body of a given mass and at rest, until it reaches a given speed. Once this point is reached, the amount of accumulated kinetic energy will remain the same unless there is a change in speed or the body returns to its state of rest by applying a force.

The kinetic energy is represented by the following expression:

Ec= ½ mv²

Where:

  • Ec is the kinetic energy, which is measured in Joules (J).
  • m is the mass measured in kilograms (kg).
  • v is the speed measured in meters over seconds (m/s).

<h3>Potential energy</h3>

On the other hand, potential energy is the energy that measures the ability of a system to perform work based on its position. In other words, this is the energy that a body has at a certain height above the ground.

Gravitational potential energy is the energy associated with the gravitational force. This will depend on the relative height of an object to some reference point, the mass, and the force of gravity.

So for an object with mass m, at height h, the expression applied to the gravitational energy of the object is:

Ep= m×g×h

Where:

  • Ep is the potential energy in joules (J).
  • m is the mass in kilograms (kg).
  • h is the height in meters (m).
  • g is the acceleration of fall in m/s².
<h3>Mechanical energy</h3>

Finally, mechanical energy is that which a body or a system obtains as a result of the speed of its movement or its specific position, and which is capable of producing mechanical work. Then:

Potential energy + kinetic energy = total mechanical energy

<h3>Principle of conservation of mechanical energy </h3>

The principle of conservation of mechanical energy indicates that the mechanical energy of a body remains constant when all the forces acting on it are conservative (a force is conservative when the work it does on a body depends only on the initial and final points and not the path taken to get from one to the other.)

Therefore, if the potential energy decreases, the kinetic energy will increase. In the same way, if the kinetics decreases, the potential energy will increase.

<h3>This case</h3>

A brick of mass 5 kg is released from rest at a height of 3 m. Then, at this height, the brick of mass has no speed, so the kinetic energy has a value of zero because it depends on the speed or moving bodies. But the potential energy is calculated as:

Ep= 5 kg× 9.8 \frac{m}{s^{2} }× 3 m

Solving:

<u><em>Ep= 147 J</em></u>

So, the mechanical energy is calculated as:

Potential energy + kinetic energy = total mechanical energy

147 J +  0 J= total mechanical energy

147 J= total mechanical energy

The principle of conservation of mechanical energy  can be applied in this case. Then, when the brick hits the ground, the mechanical energy is 147 J. In this case, considering that the height is 0 m, the potential energy is zero because this energy depends on the relative height of the object. But the object has speed, so it will have kinetic energy. Then:

Potential energy + kinetic energy = total mechanical energy

0 J +  kinetic energy= 147 J

kinetic energy= 147 J

Considering the definition of kinetic energy:

½  5 kg×v²= 147 J

v=\sqrt{\frac{2x147 J}{5 kg} }

v=7.668 m/s

Finally, when the brick hits the ground, it has a speed of 7,668 m/s.

Learn more about mechanical energy:

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brainly.com/question/11962904

#SPJ1

6 0
2 years ago
25 POINTS FOR ANSWER How are Newton’s Laws used to describe the motion of planets? Justify your response in two or more complete
Alexus [3.1K]

Pour la seule et simple raison qu'elle s'exerce entre tous les corps de l'univers ( objet, astres etc..

Si on tient compte des frottements liés aux chocs successifs des billes les une sur les autre, au bout d'un certain temps, le mouvement va cesser.

Si on dit que toute l'énergie potentielle de pesanteur est transformée en énergie cinétique, et réciproquement, donc que l'énergie mécanique est conservée au fil des chocs et des rebonds, alors, le mouvement est perpétuel. Le nombre de billes qui remontent est toujours égal au nombre de billes qu'on a lâchées.

La première loi concerne des systèmes immobiles, ou plutôt on considère des systèmes dit "isolé", c'est à dire qu'ils ne sont pas soumis à d'autre force que celle que l'on connait.

Ce qu'il faut retenir de celui ci c'est ça :

Si j'ai un système en mouvement rectiligne uniforme OU immobile, alors :

Avec F1 F2 F3, trois forces s'exercant sur mon système

Attention ! Ici je n'ai pas mit les flèches sur les différentes forces mais elles sont obligatoires ! On parle de vecteur force !

Pour la deuxième loi :

C'est le même principe, la différence c'est que l'on est en mouvement.

 

Avec a le vecteur accélération. Il y a beaucoup de ressource sur le net, n'hésite pas à regarder, la j'ai simplement pu te donner les expressions les plus connus. Mais il faudra les manipuler, et sans exercice sur lequel se baser, c'est plus difficile ! 

La troisième loi est bien moins importante que les deux autres, mais n'hésite pas à regarder sur le net, tu trouveras l'énoncé. C'est la même logique.

4 0
3 years ago
The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me
Sav [38]

Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=\dfrac{2c}{3b}.

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
  2. \rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Applying both these conditions,

\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.

For \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

For \rm t_o = \dfrac{2c}{3b},

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.

Here,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Thus, the particle reach its maximum x value at time \rm t_o = \dfrac{2c}{3b}.

7 0
3 years ago
Initial velocity=3m/s, final velocity=5m/s, displacement=2m. Find the acceleration.
shusha [124]

Answer:

Explanation:

To analyze the behavior of black bears during the night. To describe how black bears adapt to life near human. To evaluate the decline in the number of black bear.

4 0
3 years ago
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