Answer:
Refractive index of unknown liquid = 1.56
Explanation:
Using Snell's law as:
Where,
is the angle of incidence ( 65.0° )
is the angle of refraction ( 53.0° )
is the refractive index of the refraction medium (unknown liquid, n=?)
is the refractive index of the incidence medium (oil, n=1.38)
Hence,
Solving for ,
Refractive index of unknown liquid = 1.56
The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
To find the answer, we need to know about the time of flight and range of projectile motion.
<h3>What's the expression of range of a projectile motion?</h3>
- Range = U²× sin(2θ)/g
- U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
- U=√{Range×g/sin(2θ)}
- Here, range= 2.20m, = 36.5°
- U= √{2.20×9.8/sin(73)}
U= √{2.20×9.8/sin(73)} = 22.5m/s
<h3>What's the expression of time of flight in projectile motion?</h3>
- Time of flight= (2×U×sinθ)/g
- So, T= (2×22.5×sin36.5°)/9.8
= 2.73 s
Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
Learn more about the range and time period of projectile motion here:
brainly.com/question/24136952
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Answer:
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