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babunello [35]
3 years ago
11

The charge imbalance that results from this movement of charge will generate an additional electric field in the region within t

he rod In what direction will this field point? A. It will point to the right and enhance the initial applied field B. It will point to the left and oppose the initial applied field
Physics
1 answer:
deff fn [24]3 years ago
8 0

Answer:

B. It will point to the left and oppose the initial applied field

Explanation:

Let the charge be positive and  and electric field be towards right . Charge will move towards right in the field due to which positive charge will accumulate towards right and excess of positive charge will lie there . This new charge distribution will create a field towards the left which is opposite to external electric field.

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Doggo x infinity dependent on the temperature.  And the mass of mole of the gas

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1.El objetivo principal de realizar la llamada entrada en calor o calentamiento, es preparar el cuerpo para la actividad física o deportiva. Numerosas lesiones y ciertos problemas cardíacos como algunas arritmias, están asociados a la ejercitación violenta sin mediar un adecuado calentamiento.

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An electric motor has a rating of 4.0 x 10^2 watts. How much time will it take for this motor to lift a 50.-kilogram mass a vert
LenKa [72]
ΔT = ?

P = 50 kg

ΔS = 8.0 m

g = 9.8 m/s²

Pot = 4,0x10² W

Find the time :

ΔT = P * g *  ΔS / Pot

ΔT = 50 * 9.8 * 8.0 / 4.0x10²

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hope this helps!


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3 years ago
The atomic number of a nucleus increases during which nuclear reactions?
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Answer:

Option (A) : Nuclear Fusion and Beta Decay (electron emission)

3 0
3 years ago
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A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.959 g, q = 5.84 µC is locat
NemiM [27]

Answer:

8.66\times 10^{-6}\ C or 8.66\ \mu C.

Explanation:

<u>Given:</u>

  • Charge on the particle at origin = Q.
  • Mass of the moving charged particle, \rm m = 0.959\ g = 0.959\times 10^{-3}\ kg.
  • Charge on the moving charged particle, \rm q = 5.84\ \mu C = 5.84\times 10^{-6}\ C.
  • Distance of the moving charged particle from first at t = 0 time, \rm r=20.7\ cm = 0.207\ m.
  • Speed of the moving particle, \rm v = 47.9\ m/s.

For the moving particle to circular motion, the electrostatic force between the two must be balanced by the centripetal force on the moving particle.

The electrostatic force on the moving particle due to the charge Q at origin is given by Coulomb's law as:

\rm F_e = \dfrac{kqQ}{r^2}.

where, \rm k is the Coulomb's constant having value \rm 9\times 10^9\ Nm^2/C^2.

The centripetal force on the moving particle due to particle at origin is given as:

\rm F_c = \dfrac{mv^2}{r}.

For the two forces to be balanced,

\rm F_e = F_c\\\dfrac{kqQ}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow Q = \dfrac{mv^2}{r}\times \dfrac{r^2}{kq}\\=\dfrac{mv^2r}{kq}\\=\dfrac{(0.959\times 10^{-3})\times (47.9)^2\times (0.207)}{(9\times 10^9)\times (5.84\times 10^{-6})}\\=8.66\times 10^{-6}\ C\\=8.66\ \mu C.

6 0
3 years ago
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