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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The formula is=1/2(m x v^2)
so = 1/2*(0.05)*(310)^2
ans is =2402.5 joules
Answer:
D
Explanation:
The wrecking ball transfers kinetic energy to the wall.
Answer:
Part 1)
Part 2)
Part 3)
Part 4)
Part 5)
Explanation:
Part a)
Initial angular momentum of the merry go round is given as
here we know that
now we have
Part b)
Angular momentum of the person is given as
so we have
R = 1.99 m
so we have
Part 3)
Angular momentum of the person is always constant with respect to the axis of disc
so it is given as
Part 4)
By angular momentum conservation of the system we will have
Part 5)
Force required to hold the person is centripetal force which act towards the center
so we will have
If there’s no friction acting on either object then they will both be falling at a speed of 9.8m/s which is the force of gravity!