Answer:
1.06 metres per second squared
Explanation:
since friction acts against foward force
20 N - 4 N = 16 N
use Newtons 2nd law F=ma Solve for a:
a= F÷m
= 16 ÷ 15
= 1.06 metres per second squared
Answer: The question has some details missing. here is the complete question ; Point charge 1.5 μC is located at x = 0, y = 0.30 m, point charge -1.5 μC is located at x = 0 y = -0.30m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 5.0 μC at x = 0.40 m, y = 0
Explanation:
- a) First of all find the distance between the two charges;
- x = 0, y = 0.30 and x = 0.40 m, y = 0
hence, the force F = 2Kq1q2cosθ /r²...............equation 1
but cosθ = y/r = 0.3/0.5
cosθ = 0.6
plugging back to equation 1;
F = 2 x 9 x 10^9 x 1.5 x 10^-6 x 5 x 10^-6 /0.5^2
F = 540 x 10^-3
Magnitude of Force = 0.54N
b) Direction is at angle 90
The electric field between plates is 4000V/m.
An electric field (sometimes E-field) is the physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them. It also refers to the physical field for a system of charged particles.
The value of the electric field has dimensions of force per unit charge. In the metre-kilogram-second and SI systems, the appropriate units are newtons per coulomb, equivalent to volts per metre.
The voltage between points A and B is
V=E.d
E =V/d (uniform E- field only)
where d is the distance from A to B, or the distance between the plates.
Given:
distance d = 3 cm
voltage V = 120 V
Electric field E = V/d
E = 120 V / 3cm
E = 40 V / 1 cm [ 1 cm = 1/100 m ]
E = 4000 V/m.
Learn more about Electric field here:
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The correct answer is A. Installation of rigid metal conduit requires grounding and the grounding equipment used may weaken the structure.
Attenuation is the correct answer.