M/s^2 is the correct answer
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:
f3 = 102 Hz
Explanation:
To find the frequency of the sound produced by the pipe you use the following formula:
n: number of the harmonic = 3
vs: speed of sound = 340 m/s
L: length of the pipe = 2.5 m
You replace the values of n, L and vs in order to calculate the frequency:
hence, the frequency of the third overtone is 102 Hz
[Assuming that you've written 3.40 kg in 'a', and not 3.90 kg]
(a) 3,400 g x <u>0.001</u> = 3.40 kg [converting grams to kilograms]
(b) 220 cm x <u>0.01</u> = <u>2.2</u> m [converting centimeters to meters]
(c) 9.42 kg x <u>1000</u> = <u>9420</u> g [converting kilograms to grams]
(d) 6.53 m x <u>100</u> = <u>653</u> cm [converting meters to centimeters]
They both release greenhouse gases, I think
