All categories

2 years ago

Dos cargas de 1 pC estan a 1mm, la fuerza de interaccion entre dichas cargas es?

Physics

1 answer:

Naddik [55]2 years ago

6 0

Answer:

La fuerza de interacción entre dichas partículas es $8.988\times 10^{-9}\,N$ .

Explanation:

Asumamos que las dos cargas son puntuales, puesto que la fuerza de interacción es netamente electrostática, es determinada por la Ley de Coulomb:

$F = \frac{k\cdot q_{1}\cdot q_{2}}{r^{2}}$ (1)

Donde:

$k$ - Constante electrostática, medida en newton-metros cuadrados por Coulomb cuadrado.

$q_{1}$ , $q_{2}$ - Carga eléctrica, medida en Coulomb.

$r$ - Distancia entre las partículas, medida en metros.

Si sabemos que $k = 8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}$ , $q_{1} = q_{2} = 1\times 10^{-12}\,C$ y $r = 1\times 10^{-3}\,m$ , entonces la fuerza de interacción entre ambas partículas es:

$F = \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (1\times 10^{-12}\,C)^{2}}{(1\times 10^{-3}\,m)^{2}}$

$F = 8.988\times 10^{-9}\,N$

La fuerza de interacción entre dichas partículas es $8.988\times 10^{-9}\,N$ .

You might be interested in

NEED HELP ASAP

Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1} V_{o} + m_{2} U_{o}$ (2)

$p_{f}=(m_{1} + m_{2}) V_{f}$ (3)

$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

$\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2})$ (10)

Finally:

$\Delta K=-2351.25 J$ (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

6 0

2 years ago

what are the variables in an experiment that tests the distance honey flows at different temperatures? identify the independent

soldier1979 [14.2K]

Answer:

The distance that the honey flowed would be the dependent or outcome variable and the temperature of the honey would be the independent variable.

The dependent variable is what is being measured in an experiment. You can remember it by thinking “it depends on what you’re changing.”

The independent variable in an experiment is what is being changed. You can remember this by thinking “the Independent variable is what I as the scientist change.”

Explanation:

mark me brainliest plz

6 0

2 years ago

Read 2 more answers

Which of the materials would be able to scratch Quartz

Lelechka [254]

<span>anything harder than mohs scale 7 so eg Topaz, Corundum and diamond representing mohs scale 8 9 and 10 respectively.</span>

7 0

2 years ago

Which measurements are equal to 321 decimeters? Check all that apply.

S_A_V [24]