I'm not 100% sure but I think the answer is 60.4, because you multiply 3.20 by 10, to get 32 - 9 = 23, then subtract 23 from 83.4. Hope this helps you, and good luck!!!
Answer:
new atmospheric pressure is 0.9838 ×
Pa
Explanation:
given data
height = 21.6 mm = 0.0216 m
Normal atmospheric pressure = 1.013 ✕ 10^5 Pa
density of mercury = 13.6 g/cm³
to find out
atmospheric pressure
solution
we find first height of mercury when normal pressure that is
pressure p = ρ×g×h
put here value
1.013 ×
= 13.6 × 10³ × 9.81 × h
h = 0.759 m
so change in height Δh = 0.759 - 0.0216
new height H = 0.7374 m
so new pressure = ρ×g×H
put here value
new pressure = 13.6 × 10³ × 9.81 × 0.7374
atmospheric pressure = 98380.9584
so new atmospheric pressure is 0.9838 ×
Pa
Answer: 12.67 cm, 8 cm
Explanation:
Given
Normal distance of separation of eyes, d(n) = 6 cm
Distance of separation is your eyes, d(y) = 9.5 cm
Angle created during the jump, θ = 0.75°
To solve this, we use the formula,
θ = d/r, where
θ = angle created during the jump
d = separation between the eyes
r = distance from the object
θ = d/r
0.75 = 9.5 / r
r = 9.5 / 0.75
r = 12.67 cm
θ = d/r
0.75 = 6 / r
r = 6 / 0.75
r = 8 cm
Thus, the object is 12.67 cm far away in your own "unique" eyes, and just 8 cm further away to the normal person eye