<em>Force</em><em> </em><em>is</em><em> </em><em>push</em><em> </em><em>or</em><em> </em><em>pull</em><em>.</em><em>.</em><em>.</em><em>.</em>
<em>thank</em><em> </em><em>you</em>
Answer:
1/i + 1/o = 1/f thin lens equation
i = 33 * 8.9 / (33 - 8.9) = 12.2 cm to right of first lens
27 - 12.2 = 14.8 cm to left of second lens
i = 14.8 * 8.9 / (14.8 - 8.9) = 22,3 cm to right of second lens
Answer:
38.6 mi/h
Explanation:
7.4 mi/h = 7.4mi/h * (1/60)hour/min * (1/60) min/s = 0.00206 mi/s
Let v (mi/s) be your original speed, then the time t it takes to go 1 mi/s is
t = 1/v
Since you increase v by 0.00206 mi/s, your time decreases by 15 s, this means
t - 15 = 1/(v+0.00206)
We can substitute t = 1/v to solve for v

We can multiply both sides of the equation with v(v+0.00206)
v+0.00206 - 15v(v+0.00206) = v

v = -0.01278 or v = 0.01
0724 mi/s
Since v can only be positive we will pick v = 0.010724 mi/s or 0.010724*3600 = 38.6 mi/h