Rhythms that occur faster and slower than the beat are b.<span>not synchronized with the time signature. The synchronization follows the same beat or rhythm. If the time signature say is lower than the original, then the rhythm should be faster. Otherwise, the rhythm is slower than the original one.</span>
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.
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Car A will have highest speed is 83.3m/s .
<h3>What is speed ? </h3>
The rate of change of position of an object in any direction.
The S.I unit is m/s . Speed is a scalar quantity it defines only magnitude not direction
.
speed = distance /time
In case of Car A ,
We have given distance 150Km in 3 min ,
First we have convert the distance km to m
150×1000m
then conversion of min to sec
38×60sec
speed = 15000/180
speed = 83.3m/sec
In case of Car B
we have given 800m in 150 min
lets convert the time into second
150×60
Speed = 800/150×60
speed = 0.88m/ s
In case of Car C
We have given here distance 250 Km and time in 8 hours
convert km to m
25000
and time into sec
88×60×60
speed = 0.86m/ s
Hence ,Car A has highest speed amongst them .
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Lower. Water expands on lower temperatures, meaning less molecules in 1 m3, thus making it less dense