The electric field at point A located 2.00 mm above the dipole's midpoint is 5.287 X 10¹³ N/C.
<h3>
Electric field of the positive particle</h3>
The electric field is calculated as follows;
E = kq/r²
where;
- r is the distance between the charges
- k is Coulomb's constant
- q is magnitude of the charge
midpoint of 3.08 m, x = 1.54 mm
r(1.54 mm, 2.00 mm)
|r| = √(1.54² + 2²)
|r| = 2.52 mm
E = (9 x 10⁹ x 37.3 x 10⁻³)/(2.52 x 10⁻³)²
E = 5.287 X 10¹³ N/C
Thus, the electric field at point A located 2.00 mm above the dipole's midpoint is 5.287 X 10¹³ N/C.
Learn more about electric field here: brainly.com/question/14372859
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Its acceleration is 10ft/ sec
You should have the velocity as a function of time either given explicitly or implicitly (a graph)
v = ds/dt (differentiating the position vector)
integrating the acceleration.
you can use impulse or work and energy principle and also newton law of motion to find acceleration then velocity
NOT SURE IF THAT WHAT YOU WANT.
The acceleration that the same force will provide if both masses are tied together is; 6.0 m/s².
<h3>How to find the Acceleration?</h3>
We are given;
Force; F = 5 N
Acceleration of the first mass, a₁ = 8.0 m/s²
Acceleration of the second mass, a₂ = 24 m/s²
Formula for force is;
F = ma
Let us find both masses; m₁ and m₂.
m₁ = F/a₁
m₂ = F/a₂
Thus;
m₁ = 5/8 kg
m₂ = 5/24 kg
Total mass is; m = m₁ + m₂
m = 5/8 + 5/24
m = 15 + 5/24
m = 20/24 kg
Thus, acceleration if they are both tied together is;
a = F/m
a = 5/(20/24)
a = 6.0 m/s².
Read more about Acceleration at; brainly.com/question/605631
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