Answer:
a. 0.8 cm
Explanation:
The distance of the object from the lens, u = 1 cm
The magnification of the lens, m = 5
The focus of a lens formula is given as follows;
The magnification of the lens, m = -v/u
Where;
v = The distance of the image from the lens
Therefore, we have;
v = m × u
∴ v = 5 × 1 cm = 5 cm (on the other side of the lens)
From which we get;
The focal length ≈ 0.8 cm
Answer:
his acceleration rate is -0.00186 m/s²
Explanation:
Given;
initial position of the car, x₀ = 100 miles = 160, 900 m ( 1 mile = 1609 m)
time of motion, t₀ = 60 minutes = 60 mins x 60 s = 3,600 s
final position of the car, x₁ = 150 miles = 241,350 m
time of motion, t₁ = 100 minutes = 100 mins x 60 s = 6,000 s
The initial velocity is calculated as;
u = 160, 900 m / 3,600 s
u = 44.694 m/s
The final velocity is calculated as;
v = 241,350 m / 6,000 s
v = 40.225 m/s
The acceleration is calculated as;
Therefore, his acceleration rate is -0.00186 m/s²