Answer:
a. 7.62cm
b. Real and inverted
c. 2.76 cm
d. 3450
Explanation:
We proceed as follows;
a. the lens equation that relates the object distance to the image distance with the focal length is given as follows;
1/f = 1/p + 1/q
making q the subject of the formula;
q = pf/p-f
From the question;
p = 4.70m
f = 7.5cm = 0.075m
Substituting these, we have ;
q = (4.7)(0.075)/(4.7-0.075) = 0.3525/4.625 = 0.0762 = 7.62 cm
b. The image is real and inverted since the image distance is positive
c. We want to calculate how tall the image is
Mathematically;
h1 = (q/p)h0
h1 = (7.62/4.70)* 1.7
h1 = 2.76 cm
d. We want to calculate the number of pixels that fit into this image
Mathematically:
n = h1/8 micro meter
n = 2.76cm/8 micro meter = 2.76 * 10^-2/8 * 10^-6 = 3450
Answer:
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Answer:
269 m
45 m/s
-58.6 m/s
Explanation:
Part 1
First, find the time it takes for the package to land. Take the upward direction to be positive.
Given (in the y direction):
Δy = -175 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
(-175 m) = (0 m/s) t + ½ (-9.8 m/s²) t²
t = 5.98 s
Next, find the horizontal distance traveled in that time:
Given (in the x direction):
v₀ = 45 m/s
a = 0 m/s²
t = 5.98 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (45 m/s) (5.98 s) + ½ (0 m/s²) (5.98 s)²
Δx = 269 m
Part 2
Given (in the x direction):
v₀ = 45 m/s
a = 0 m/s²
t = 5.98 s
Find: v
v = at + v₀
v = (0 m/s²) (5.98 s) + (45 m/s
v = 45 m/s
Part 3
Given (in the y direction):
Δy = -175 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: v
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (-9.8 m/s²) (-175 m)
v = -58.6 m/s