Answer:
Option 3: -48 cm
Explanation:
We are given:
refractive index; n = 1.5
radius of curvature; r2 = 24 cm
Formula for the focal length is given as;
1/f = (n - 1) × [(1/r1) - (1/r2)]
As r1 tends to infinity, 1/r1 = 0
Thus,we now have;
1/f = (n - 1) × (-1/r2)
Plugging in the relevant values;
1/f = (1.5 - 1) × (-1/24)
1/f = -0.02083333333
f = -1/0.02083333333
f = -48 cm
Answer:
last option is the correct one
The speed of tsunami is a.0.32 km.
Steps involved :
The equation s = 356d models the maximum speed that a tsunami can move at. It reads as follows: s = 200 km/h d =?
Let's now change s to s in the equation to determine d: s = 356√d 200 = 356√d √d = 200 ÷ 356 √d = 0.562 Let's square the equation now by squaring both sides: (√d)² = (0.562) ² d = (0.562)² = 0.316 ≈ 0.32
As a result, 0.32 km is roughly the depth (d) of water for a tsunami moving at 200 km/h.
To learn more about tsunami refer : brainly.com/question/11687903
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Answer:
a. Acceleration, a = 1.88 m/s²
b. Time, t = 7.87 seconds.
Explanation:
Given the following data;
Initial velocity, U = 14.5m/s
Final velocity, V = 29.3m/s
Distance, S = 172m
a. To find the acceleration of the speedboat;
We would use the third equation of motion;
V² = U² + 2aS
Substituting into the formula
29.3² = 14.5² + 2a*172
858.49 = 210.25 + 344a
344a = 858.49 - 210.25
344a = 648.24
a = 648.24/344
Acceleration, a = 1.88 m/s²
b. To find the time;
We would use the first equation of motion;
V = U + at
29.3 = 14.5 + 1.88t
1.88t = 29.3 - 14.5
1.88t = 14.8
Time, t = 14.8/1.88
Time, t = 7.87 seconds.
