The magnitude of the electric field at x =4m on the x axis at this time 1 N/C.
<h3>Electric field at position 4 m</h3>
Electric field at a given distance is calculated as follows;
E = kq/r²
E₂ = (9 x 10⁹ x q)/(2²)
E₂ = 2.25 x 10⁹q
E₂ + E₀ = 0
2.25 x 10⁹q + 4 = 0
2.25 x 10⁹q = - 4
q = -4 / (2.25 x 10⁹)
q = -1.78 x 10⁻⁹
E₄ = (9 x 10⁹ x (-1.78 x 10⁻⁹) ) / (4²)
E₄ = - 1 N/C
|E₄| = 1 N/C
Thus, the magnitude of the electric field at x =4m on the x axis at this time 1 N/C.
The complete question is below:
Suppose a uniform electric field of 4N/C is in the positive x direction. When a charge is placed and at a fixed to the origin, the resulting electric field on the x axis at x =2m becomes zero. What is the magnitude of the electric field at x =4m on the x axis at this time?
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Answer:
9m^3
Explanation:
Given data
volume v1= 3m^3
volume v2= ???
Temperature T1= 20.0°C.
Temperature T2= 60.0°C.
Applying the relation for temperature and volume
V1/T1= V2/T2
substitute
3/20= V2/60
3*60= V2*20
180= 20*V2
180/20= V2
V2= 9m^3
Hence the final volume is 9m^3
Kepler discovered the elliptical shape in the orbital path.<span />
French Physicist Leon Foucault introduced his pendulum in 1851.
By that time, it was generally accepted that the Earth rotates, but
Foucault's pendulum was the first simple demonstration that made
the Earth's <u>rotation</u> easy to prove.
The traveler's total displacement is 53.2 km.
The shortest distance between a particle's starting position and destination is referred to as its displacement vector. Displacement has a distinct physical magnitude than distance. Let's use the motion of a particle making a complete circle as an example. The displacement in this motion is zero, yet the distance traveled is equal to the circle's circumference. The individual displacement vectors are as follows: d1
=22^ikm→d2
= (29.7cos45i29.7sin45) kmd3=104jkm
The traveler's complete disadvantage is,
→d=→d1+→d2+→d3→d
=22^ikM+29.7
cos45^−29.7sin45o) km−10.4^j km→
d=(43^i−31.4^j) km
The magnitude of the total displacement is,
d=√432+31.4/2Km d≈53.2km
The direction is,
θ=tan−1(−31.4/43)
θ=−36.1or 36.1 south of east.
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