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:
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Answer:
The answer is mutualism because they are both on the receiving and giving ends
Respiratory system.
Oversimplified Explanation: they enter the lungs, which is part of the respiratory system.
Answer:
<h2>80,000 J</h2>
Explanation:
The potential energy of a body can be found by using the formula
PE = mgh
where
m is the mass
h is the height
g is the acceleration due to gravity which is 10 m/s²
From the question we have
PE = 250 × 10 × 32
We have the final answer as
<h3>80,000 J</h3>
Hope this helps you
Given: A cubic tank holds 1,000.0 kg of water.
Mass of water in tank (m) = 1000.0 kg
Density of water (d) = 1000.0 kg /m³
Concept: Volume(V) = Mass / Density
Since the tank holds these water in it so the volume of water will be equal to the volume of the tank.
Hence, the volume of the tank = Mass of water / Density of water
or, = 1000.0 kg / 1000.0 kg m⁻³
or, = 1.0 m³
Since tank is cubical in shape. Let its side be 'x'
The volume of tank (x³) = 1.0 m³
or. side of tank (x) = 1.0 m
Hence, the dimensions of the tank will be 1.0 m.
