Answer:
what should be done io protect forests
Answer:
(a) 8.362 rad/sec
(b) 6.815 m/sec
(c) 9.446 
(d) 396.22 revolution
Explanation:
We have given that diameter d = 1.63 m
So radius 
Angular speed N = 79.9 rev/min
(a) We know that angular speed in radian per sec

(b) We know that linear speed is given by

(c) We have given final angular velocity 
And 
Time t = 63 sec
Angular acceleration is given by 
(d) Change in angle is given by

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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The solution for this problem is:
r = [(2.90 + 0.0900t²) i - 0.0150t³ j] m/s²
this is for t in seconds and r in meters
v = dr/dt = [0.180t i - 0.0450t² j] m/s²
tan(-36.0º) = -0.0450t² / 0.180t
0.7265 = 0.25t
t = 2.91 s is the velocity vector of the insect