Answer:
(a)

(b) 

Explanation:
Let us take the north direction to be the positive y-axis and the east to be positive x-axis.
First day:
25.0 km southeast, which implies
south of east. The y-component will be negative and the x-component will be positive.


Second day:
She starts off at the stopping point of last day. This time, both the y- and x-components are positive.


Therefore, total displacements:


Magnitude of displacements,

Direction,

Answer:
120 m/s if m/s means miles per second than
Explanation:
15 × 8
Answer:
A. 30.7cm
B. 
C. The electric field is directed away from the point of charge
Explanation:
A.

B.
Considering Gauss's law

C. The electric field directed away from the point of charge when the charge is positive.
Answer:
1) q₁ = 12.987 cm
, b) L = 17.987 cm
, c) m = 179.87
Explanation:
We can solve the geometric optics exercises with the equation of the constructor
1 / f = 1 / p + 1 / q
where f is the focal length, p and q are the distance to the object and the image respectively.
Let's apply this equation to our case
1) f = 5mm = 0.5 cm
p₁ = 5.2 mm = 0.52 cm
h = 0.1 mm = 0.01 cm
1 / q₁ = 1 / f- 1 / p
1 / q₁ = 1 / 0.5 - 1 / 0.52 = 2 - 1.923
1 / q₁ = 0.077
q₁ = 12.987 cm
2) in this part they tell us that the eyepiece creates an image at infinity, therefore the object that comes from being at the focal length of the eyepiece
p₂ = 5 cm
The absolute thing that goes through the two lenses is
L = q₁ + p₂
L = 12.987 +5
L = 17.987 cm
3) This lens configuration forms the so-called microscope, whose expression for the magnifications
m = -L / f_target 25 cm / f_ocular
m = - 17.987 / 0.5 25 / 5.0
m = 179.87