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densk [106]
2 years ago
4

for the same charge q, the electric field strength measured at point x is the electric field measured at point y. the distance o

f point x from the charge is how many times the distance of point y from the charge?
Physics
1 answer:
scZoUnD [109]2 years ago
7 0

Due to a single charge and the additional distance =2r, the electric field is E r 2.

New intensity is consequently equal to E/4.

<h3>Where is electric field?</h3>

From the point charge, the electric field radiates outward in all directions. Spherical equipotential surfaces are represented by the circles. The vector sum of the individual fields may be used to calculate the electric field from any number of point charges.

Volts per metre (V/m) is the electrical field's SI unit. The Newton's per coulomb unit is the same as this one. These units are derivations from Newton, which stands for force, and Coulomb, which stands for charge.

Being a vector quantity, the electric. We are aware that the electric field is defined as the force to test charge ratio.

learn more about electric field refer

brainly.com/question/24317284

#SPJ4

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Una fuerza F de 200 lb actúa a lo largo de AB, sobre la rampa mostrada. fuerza de F respecto del eje OC. Calcule el momento de f
ruslelena [56]

Answer:

Moc = -613.25 [lb*in]

Explanation:

Este problema se puede resolver mediante la mecánica vectorial, es decir se realizara un analisis de vectores.

Primero se calculara el momento de la fuerza F_AB con respecto al punto O, debemos recordar que el momento con respecto a un punto se define como el producto cruz de la distancia por la fuerza.

M_{o}=r_{A/O} * F_{AB} (producto cruz)

Necesitamos identificar los puntos:

O (0,0,0) [in]

A (12,0,0) [in]

B (0, 24,8) [in]

C (12,24,0) [in]

r_{A/O}=(12,0,0) - (0,0,0)\\r_{A/O} = 12 i + 0j+0k [in]\\AB = (0,24,8) - (12,0,0)\\AB = -12i+24j+8k [in]\\[LAB]=\frac{-12i+24j+8k}{\sqrt{(12)^{2} +(24)^{2} +(8)^{2} } }\\ LAB=-\frac{3}{7} i+\frac{6}{7}j+\frac{2}{7}k

El ultimo vector calculado corresponde al vector unitario (magnitud = 1) de AB. El vector fuerza corresponderá al producto del vector unitario por la magnitud de la fuerza = 200 [lb].

F_{AB}=-\frac{600}{7} i +\frac{1200}{7}j+\frac{400}{7} k [Lb]

De esta manera realizando el producto cruz tenemos

M_{O}=r_{A/O} * F_{AB}

M_{O}=0i-685.7j+2057.1k [Lb*in]

Para calcular el momento con respecto a la diagonal OC, necesitamos el vector unitario de esta diagonal.

OC = (12,24,0)-(0,0,0)\\OC= 12i+24j+0k[Lb]\\LOC = \frac{12i+24j+0k}{\sqrt{(12)^{2} +(24)^{2} +(0)^{2} } } \\LOC=\frac{12}{\sqrt{720}}i+\frac{24}{\sqrt{720}}j  +0k

El vector con respecto al eje OC, es igual al producto punto del momento en el punto O por el vector unitario LOC

M_{OC}=L_{OC}*M_{O}\\M_{OC}=(\frac{12}{\sqrt{720}}i +\frac{24}{\sqrt{720}} j+0k )* (0i-685.7j+2057.1k)\\M_{OC}= -613.32[Lb*in]

7 0
3 years ago
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