I think its either C or D. I tried, couldn't figure the last part out. Hope this helped though!! Have a great day! :D
William Gilbert is known as the father of electricity.
Answer:
A) q = -8.488 cm
, B) m = 0.29
Explanation:
A) For this exercise in geometric optics, we will use the equation of the constructor
where p and q are the distance to the object and image, respectively and f is the focal length
in our case the distance the object is p = 29 cm the focal length of a diverging lens is negative and indicates that it is f = - 12 cm

we calculate

= - 0.1178
q = -8.488 cm
the negative sign indicates that the image is virtual
B) the magnification is given

we substitute
m =
m = 0.29
the positive sign indicates that the image is right
Answer:
a. v₁ = 16.2 m/s
b. μ = 0.251
Explanation:
Given:
θ = 15 ° , r = 100 m , v₂ = 15.0 km / h
a.
To determine v₁ to take a 100 m radius curve banked at 15 °
tan θ = v₁² / r * g
v₁ = √ r * g * tan θ
v₁ = √ 100 m * 9.8 m/s² * tan 15° = 16.2 m/s
b.
To determine μ friction needed for a frightened
v₂ = 15.0 km / h * 1000 m / 1 km * 1h / 60 minute * 1 minute / 60 seg
v₂ = 4.2 m/s
fk = μ * m * g
a₁ = v₁² / r = 16.2 ² / 100 m = 2.63 m/s²
a₂ = v₂² / r = 4.2 ² / 100 m = 0.18 m/s²
F₁ = m * a₁ , F₂ = m * a₂
fk = F₁ - F₂ ⇒ μ * m * g = m * ( a₁ - a₂)
μ * g = a₁ - a₂ ⇒ μ = a₁ - a₂ / g
μ = [ 2.63 m/s² - 0.18 m/s² ] / (9.8 m/s²)
μ = 0.251