It is the energy of the electrons I believe
Answer:
the tension in the part of the cord attached to the textbook is 7.4989 N
Explanation:
Given the data in the question;
As illustrated in the image below;
first we determine the value of the acceleration,
along vertical direction; we use the second equation of motion;
y = ut + at²
we substitute;
0 m/s for u, 1.29 m for y, 0.850 s for t,
1.29 = 0×0.850 + ×a×(0.850)²
1.29 = 0.36125a
a = 1.29 / 0.36125
a = 3.5709 m/s²
Now when the text book is moving with acceleration , the dynamic equation will be;
T₁ = m₁a
where m₁ is the mass of the text book ( 2.10 kg )
a is the vertical acceleration ( 3.5709 m/s² )
so we substitute
T₁ = 2.10 × 3.5709
T₁ = 7.4989 N
Therefore, the tension in the part of the cord attached to the textbook is 7.4989 N
Osmosis, Oxygen, Orbit, Organism,
Q = magnitude of charge at each of the three locations A, B and C = 2 x 10⁻⁶ C
r₁ = distance of charge at origin from charge at B = 50 - 0 = 50 cm = 0.50 m
r₂ = distance of charge at origin from charge at C = 100 - 0 = 100 cm = 1 m
F₁ = magnitude of force by charge at B on charge at origin
F₂ = magnitude of force by charge at C on charge at origin
Magnitude of force by charge at B on charge at origin
inserting the values
F₁ = 0.144 N
Magnitude of force by charge at C on charge at origin
inserting the values
F₂ = 0.036 N
Net force on the charge at the origin is given as
F = F₁ + F₂
F = 0.144 + 0.036
F = 0.18 N
from the diagram , direction of net force is towards left or negative x-direction.